Part IV · Quantum Mechanics — Chapter 4.8

The Oscillator, and the Ladder

One potential, solved twice over. The differential equation was done in Chapter 4.5 and this chapter does not repeat it. What it does instead is take the same operator apart into two factors, discover that the factorisation fails by exactly one commutator, and get the whole spectrum out of that failure. The answer is famous. The method is what Parts V and VII are made of.

Tools you'll need  — Chapter 4.6 §4.2 for the identification of $\hat H$ with $\hat p^{2}/2m+V(\hat x)$, §5 for $\hat p=-\ii\hbar\,\dd/\dd x$ in the position representation, §9 for stationary states, and §10 for a packet that spreads. Chapter 4.5 §5 for the Hermite functions, their generating function and their completeness, §5.5 for which operator $\hat H_{\text{osc}}$ actually is, and §8.2 for when an expansion in eigenstates is legitimate. Chapter 4.4 §3 for an operator as a formula together with a domain. Chapter 4.2 §8 for the canonical commutator, and §5.3 for the spread of an observable in a state. Chapter 4.7 §2 for parity and §3.5 for a zero-point energy read as an uncertainty. Chapter 1.3 §4.4 for $\oint p\,\dd q$ and the Bohr–Sommerfeld guess, and §6.3 for the algebra a commutator obeys. Chapter 1.1's Problem 2 for the reduced mass. Chapter 0.9 §6 for the bandwidth theorem and §6.5 for the functions that saturate it. Chapter 0.8 §2.1 for a first-order linear equation solved by separation, §3.1 for the dimension of a solution space, §4.2 for the classical complex amplitude, and §4.4 for the phase-space ellipse and its area. Chapter 0.5 §1.1 for the inner-product axioms and §4 for the adjoint. Chapter 0.2 §4 for the Gaussian integral.

1 · Two routes, and why only one is taken

Here is where this section is going. There are exactly two standard ways to solve the quantum oscillator, one of them is already in this book, and this chapter takes the other. By the end of this section you will know what the route not taken looks like, why skipping it costs nothing, and what the route taken is for, which is not the answer but the technique.

1.1 · The operator, and what is already settled about it

Chapter 4.6 §4.2 identified the Hamiltonian of a particle in a potential as $\hat p^{2}/2m+V(\hat x)$, and putting the harmonic potential into it gives the operator this chapter is about:

Chapter 4.4 spent a chapter on the fact that a formula like this is not yet an operator, because an operator is a formula together with a domain, and that for the particle in a box the missing information was four real numbers. Here that question is already answered, and the answer is not something this chapter has to work for. Chapter 4.5 §5.5 exhibited a complete orthonormal family of eigenfunctions for (4.8.1) and used it to write the operator as multiplication by a real function in a different description, which makes it self-adjoint by construction. That section said in so many words that it was settling the question in order to save an argument here. So we may treat (4.8.1) as a genuine observable and get straight to its spectrum.

What that means has a name, and two earlier chapters gave it. Chapter 0.5's closing list said the spectral theorem would return in this part of the book as the statement that solving a system means diagonalising its Hamiltonian, and Chapter 4.2's closing list picked out this chapter as one of the places it happens. Everything below is that diagonalisation: a complete orthonormal family of eigenvectors of (4.8.1), together with the number attached to each. Section 6.4 then does the same job a second time on a computer, and the two answers are compared.

1.2 · The route already taken

The first route is the differential equation. Write (4.8.1) in the position representation, strip out the constants with the length $x_{0}=\sqrt{\hbar/m\omega}$, and the eigenvalue problem becomes $-u''+\xi^{2}u=\varepsilon u$ in the dimensionless variable $\xi=x/x_{0}$, with $\varepsilon=2E/\hbar\omega$ the energy measured in half-quanta. Peel off the Gaussian that the large-$\xi$ behaviour demands, write what is left as a power series, and the equation turns into a recurrence between successive coefficients. The series then has to terminate, because if it does not the coefficients eventually behave like those of $\ee^{+\xi^{2}}$ and the solution is not normalisable. Termination happens only at particular $\varepsilon$, and those are the eigenvalues.

That route exists, it is standard, and it is not being taken here. Two reasons, and the first is short. The series route does two separate jobs, and this book has both of them covered elsewhere. Exhibiting the solutions is Chapter 4.5 §5's job, and it is done. That section defined the Hermite polynomials by a generating function and extracted their recurrences. It then showed that the resulting functions satisfy the equation above and read the eigenvalues off, without a single power series. Excluding every other energy is the second job, the one termination performs in the series argument, and §4 below does it with a norm instead. The guarantee that no state has been missed is Chapter 4.5 §5.4's completeness proof, which §5.5 collects. Doing a Frobenius expansion here would produce the same number from the same side of the problem, and this book does not have a theory of special functions to spend it on.

The second reason is the one that matters. The algebraic route is not a cheaper way to the same answer. It is a different technique, and the technique is what the rest of the book runs on. Chapter 5.3 quantises a field by attaching a copy of this chapter's algebra to every mode, and the integer that labels a rung becomes the number of particles present. Chapter 7.4 does the same for a vibrating string, where the rungs are the particle spectrum. Neither of those uses a Hermite polynomial anywhere. What they use is $[\hat a,\hat a^{\dagger}]=1$ and what follows from it, and that is what §§2 to 4 build. This book builds every special function it needs algebraically and never by a series, here and in Chapters 4.12 and 4.13 alike, and the decision was taken once for all three.

1.3 · What Chapter 4.7 hands over

Three things were promised to this chapter by name. Here they are, collected before starting rather than after finishing.

• The fifth exactly solvable potential. Chapter 4.7 solved four and said this is the one its method cannot reach, since the potential is nowhere constant and there are no pieces to join. What replaces matching is an algebra, and §2 is where the replacement happens.
• One of exactly two routes onward. Once the potential stops being piecewise constant you either solve one case exactly by algebra or approximate all cases. Chapter 4.10 does the second, and this chapter's exact answer is what its approximation gets scored against.
• The alternating parities, by a third route. Chapter 4.7 found that the states of a symmetric potential alternate even and odd up the ladder, and got it two ways: from a non-degeneracy argument in its §2.4, and from the roots of a matching condition in its §4.3. Section 5.6 below gets the same alternation for the oscillator out of the algebra alone, without looking at a single wavefunction.

What this chapter needs from Chapter 4.7 is only the shape of the answer: confinement gives a discrete ladder, the parities alternate, and the ground state does not sit at the bottom of the well. All three turn up again here, and none of them is assumed.

2 · Factorising $\hat x^{2}+\hat p^{2}$, and what the leftover is

Here is where this section is going. We are going to try to write the energy as a product rather than a sum, because a product of two first-order objects is a far easier thing to handle than a second-order differential operator. The attempt does not quite work, and the interesting part is the amount by which it fails. That amount is the commutator of position and momentum, it is a pure number, and it is the ground-state energy of the oscillator arriving before any eigenvalue problem has been solved.

2.1 · One length, and the operator with the constants removed

Constants are in the way, and there is exactly one length that removes them. Out of $\hbar$, $m$ and $\omega$ you can build precisely one quantity with the dimensions of length, and one with the dimensions of momentum:

Chapter 4.5 §5.1 introduced the first of these for the same reason. Now define dimensionless operators by measuring against them, $\hat X=\hat x/x_{0}$ and $\hat P=\hat p/p_{0}$. Two things happen at once. The Hamiltonian loses its constants, since $\hat p^{2}/2m=\half\hbar\omega\hat P^{2}$ and $\half m\omega^{2}\hat x^{2}=\half\hbar\omega\hat X^{2}$, and the canonical commutator of Chapter 4.2 §8 loses its $\hbar$, because dividing $[\hat x,\hat p]=\ii\hbar$ by $x_{0}p_{0}=\hbar$ leaves

Everything in this chapter comes out of those two statements. The second one is Chapter 4.2's postulate P6, the one relation about position and momentum that was assumed rather than derived, wearing different units. Nothing else about the operators will be used.

2.2 · The factorisation, and the piece that will not go

We want the energy as a product, and the shape of $\hat X^{2}+\hat P^{2}$ says what product to try. For two ordinary numbers, $X^{2}+P^{2}=(X-\ii P)(X+\ii P)$, and the reason that identity works is that the two cross terms $+\ii XP$ and $-\ii PX$ cancel. So write the same product with operators and watch the cancellation fail:

The cross terms do not cancel. What is left over in (4.8.4) is $+\ii\big[\hat X,\hat P\big]$, and (4.8.3) makes that $\ii\cdot\ii=-1$. Rearranging (4.8.4) to put the energy on the left carries that term across the equals sign and flips its sign, so the term standing beside the product below is $-\ii[\hat X,\hat P]$, which is $+1$:

The sum of two squares has been written as a product, and the price of doing so is one extra term. That term is not an approximation, not a correction and not a small quantity. It is the commutator itself, and it is there for exactly the reason that position and momentum do not commute. Multiply (4.8.5) by $\hbar\omega/2$ and read what the price is in units of energy:

The $\half\hbar\omega$ is already here. It has arrived before any eigenvalue has been computed, before any wavefunction has been written down, and before any boundary condition has been imposed. It is the arithmetic cost of factorising a sum of two squares whose factors do not commute. Anything that later calls itself the zero-point energy of this oscillator is this term, and it can no more be removed than the commutator can. Two things are still owed before the term can be called the bottom of the spectrum rather than an offset above which the levels might start anywhere. One is that the product standing beside it in (4.8.6) is never negative, which is §4.2. The other is that a state the right-hand factor sends to zero exists at all, which is §5.2. Both are paid inside this chapter, and neither is assumed here.

2.3 · Naming the two factors

The two factors are going to be used constantly, so give them names, and choose the numerical factor in front so that the algebra afterwards is as clean as possible. Define

The second form is the first one with (4.8.2) substituted back, and it is the form to use when a real mass and a real frequency are in play. The factor $1/\sqrt2$ is a convention and §2.4 is where it earns its keep.

Now the other factor. Chapter 0.5 §4 defined the adjoint by how it moves across the inner product, and one of the three rules it derived on the spot is that $(\alpha\hat A)^{\dagger}=\bar\alpha\hat A^{\dagger}$. Both $\hat x$ and $\hat p$ are self-adjoint, so taking the adjoint of $\hat X+\ii\hat P$ leaves the two operators alone and conjugates the $\ii$:

These two are the only new objects in the chapter's first half, and inverting the pair gives position and momentum back whenever they are wanted:

2.4 · The one relation everything else comes from

With the factor of $1/\sqrt2$ in place, the commutator of the two new objects is as simple as it can be. Expanding $[\hat a,\hat a^{\dagger}]$ using (4.8.7) and (4.8.8), the $\hat X$ with $\hat X$ and $\hat P$ with $\hat P$ terms commute and drop out, and what survives is

That is what the $1/\sqrt2$ bought: without it the answer would be $2$, and every square root later in the chapter would carry a stray factor. And rewriting (4.8.6) with the new names, since $\hat a^{\dagger}\hat a=\half(\hat X-\ii\hat P)(\hat X+\ii\hat P)$,

What has happened is a genuine change of problem. Equation (4.8.1) was a second-order differential operator and its eigenvalue problem was a second-order differential equation. Equation (4.8.11) is a product of two first-order objects and a constant, and the only property of those objects that the next two sections use is (4.8.10). Not their form, not the space they act on, not what $\hat x$ and $\hat p$ mean. One line of algebra.

3 · The ladder

Here is where this section is going. We have a Hamiltonian built out of two objects satisfying one commutation relation. What we want from them is motion between eigenstates: an operation that takes a state of definite energy and hands back another one, higher up. That operation exists, one commutator proves it in two lines, and the manoeuvre that does it runs three more times before this book ends.

3.1 · The number operator

The combination that appears in (4.8.11) deserves its own name, because every question about the energy is now a question about it. Define

Its adjoint is $(\hat a^{\dagger}\hat a)^{\dagger}=\hat a^{\dagger}\hat a$, using Chapter 0.5 §4's rule that taking the adjoint of a product reverses it, so $\hat N$ is symmetric. And because $\hat H$ is $\hat N$ multiplied by a positive constant and shifted by another, the two have exactly the same eigenvectors, with eigenvalues related by $E=(\nu+\half)\hbar\omega$. So the entire problem is now to find the eigenvalues of $\hat N$. We do not yet know that they are integers, and it would be circular to call them $n$ before showing it, so write $\nu$ for a generic eigenvalue and let the algebra decide what $\nu$ can be.

3.2 · Two commutators

To move between eigenstates of $\hat N$ we need to know how $\hat N$ reacts when $\hat a^{\dagger}$ is pushed past it, which is exactly what a commutator records. Compute it from (4.8.10) alone. Since $\hat a^{\dagger}$ commutes with itself, the only thing to move is the $\hat a$:

Neither line needs an identity quoted from anywhere. Writing the first one out, $\hat a^{\dagger}\hat a\hat a^{\dagger}-\hat a^{\dagger}\hat a^{\dagger}\hat a$ has a common factor of $\hat a^{\dagger}$ on the left, and what remains inside is $\hat a\hat a^{\dagger}-\hat a^{\dagger}\hat a$, which is (4.8.10). The second goes the same way with the common factor on the right. Chapter 1.3 §6.3 noted that a commutator obeys the Leibniz rule exactly as a Poisson bracket does, and this is that rule used once. The two relations differ only in sign, so it is enough to follow the first and read the second off at the end.

3.3 · What a commutator like that does

A commutator is a difference of two orderings, and the use of one is that it lets you move an operator through another at the cost of a known remainder. Here the remainder is the operator itself, and that is what makes the relation powerful. Suppose $\ket\nu$ is an eigenvector of $\hat N$ with eigenvalue $\nu$. Ask what $\hat N$ does to the vector $\hat a^{\dagger}\ket\nu$, and use (4.8.13) to move $\hat N$ past $\hat a^{\dagger}$ so that it lands on the eigenvector, where we know what it does:

That line is the whole of the method. The vector $\hat a^{\dagger}\ket\nu$ is again an eigenvector of $\hat N$, and its eigenvalue is one greater. Running the same two lines with $[\hat N,\hat a]=-\hat a$ gives the mirror statement, that $\hat a\ket\nu$ has eigenvalue $\nu-1$. So the two operators move a state up and down a ladder of eigenvalues spaced exactly one apart, which in energy is a ladder spaced $\hbar\omega$ apart.

One qualification has to be attached now rather than later, because §4 turns on it. Equation (4.8.14) is an identity between vectors, and the zero vector satisfies it as readily as any other. So the honest reading is that $\hat a^{\dagger}\ket\nu$ is either an eigenvector with eigenvalue $\nu+1$ or the zero vector, which is not a state at all. The same disjunction holds downward. Everything interesting about the oscillator's spectrum comes from asking when the second alternative happens.

3.4 · The move, named once, because it happens three more times

Nothing in §3.3 used the oscillator. Strip the physics out and what remains is a pattern with three ingredients: an operator $\hat N$ whose eigenvalues you want, a second operator $\hat A$, and a commutation relation $[\hat N,\hat A]=\lambda\hat A$ for some number $\lambda$. Whenever those three hold, $\hat A$ carries eigenvectors of $\hat N$ to eigenvectors of $\hat N$ with the eigenvalue shifted by $\lambda$, and the proof is (4.8.14) with the letters changed. Call it a commutator that shifts an eigenvalue. It runs three more times in this book, and each time the algebra around it is heavier and the shape harder to see, so here are the three in advance.

• Chapter 4.11 §4 runs it on angular momentum. There $\hat N$ is $\hat J_{z}$, the shifting operators are $\hat J_{\pm}=\hat J_{x}\pm\ii\hat J_{y}$, and the relation is $[\hat J_{z},\hat J_{\pm}]=\pm\hbar\hat J_{\pm}$. The extra work in that chapter is that the ladder has to close at both ends rather than one, and the argument that closes it is §4 below run twice.
• Chapter 4.13 §4 runs it on hydrogen. There the shifting operator carries you between different values of the orbital angular momentum at fixed energy, and it is built by factorising the radial Hamiltonian the way §2 factorised this one. It is the same manoeuvre, and that chapter says so.
• Chapter 7.3 builds the Virasoro algebra out of oscillator commutators, and Chapter 7.4 quantises the string with it, where the operator counting rungs is the one whose eigenvalue fixes a particle's mass. The algebra there is this chapter's with an index running over the string's modes.

So the move appears twice more inside the next five chapters and once more in Part VII. The reason to name it here, on the smallest example there is, is that here the whole content is (4.8.10) and a two-line calculation. By Chapter 4.11 there are three operators, a Casimir and a sign convention in the way, and a reader who has not been told what to look for will read the algebra instead of the shape.

4 · Why it stops

Here is where this section is going. Section 3 built a machine that lowers the eigenvalue by one every time it is applied, and nothing so far stops it running forever into negative numbers. Something must, and the something is not a physical argument, an appeal to a boundary condition, or a statement that the energy ought to be positive. It is that a length cannot be negative. By the end of this section the spectrum will be in hand, derived from (4.8.10) and one axiom of the inner product.

4.1 · The problem, stated sharply

Suppose $\hat N\ket\nu=\nu\ket\nu$ with $\ket\nu$ a genuine non-zero state. Apply $\hat a$ once and, by §3.3, you have a state with eigenvalue $\nu-1$ or the zero vector. Apply it $k$ times and you have a state with eigenvalue $\nu-k$ or the zero vector. If the zero vector never turns up, then choosing $k$ larger than $\nu$ produces an eigenvector of $\hat N$ with a negative eigenvalue, and choosing $k$ larger still produces one as negative as you please. Nothing in §3 forbids that. The spectrum would run down without end, the energies with it, and the oscillator would have no ground state.

So the question is not whether the ladder terminates. The question is what could possibly terminate it, and most books answer with a sentence to the effect that it must stop somewhere. It does not have to stop somewhere. It stops because of the following.

4.2 · A norm is not negative, and that is the whole argument

We want a lower bound on $\nu$, so look for a quantity that is manifestly non-negative and that has $\hat N$ in it. There is one, and it is the length of the vector $\hat a\ket\psi$. Chapter 0.5 §1.1 laid down the axioms of an inner product, and the last of them is that $\avg{v,v}\ge0$ for every vector, with equality only for the zero vector. That is not a theorem about quantum mechanics. It is part of what the word length means. Apply it to $\hat a\ket\psi$ and move one factor of $\hat a$ across the inner product using Chapter 0.5 §4's defining property of the adjoint:

The middle step is the only one carrying a condition, and the condition is the domain question flagged in §2.4: it needs $\ket\psi$ to be in the domain of $\hat a$ and $\hat a\ket\psi$ in the domain of $\hat a^{\dagger}$. For the vectors this section is about, the eigenvectors of $\hat N$, both hold automatically. The operator $\hat N=\hat a^{\dagger}\hat a$ is a product, and the domain of a product is by definition the set of vectors the two factors can be applied to in turn. That is what Chapter 4.4 §3 means by an operator being a formula together with a domain. Calling $\ket\nu$ an eigenvector of $\hat N$ already says that $\ket\nu$ lies in the domain of $\hat N$, so it is one of those vectors, and nothing about which $\nu$ it belongs to has entered. That matters for §4.3, which applies (4.8.15) to a hypothetical eigenvector that is deliberately not one of §5's states. Section 5's states satisfy both conditions as well, being polynomials times a Gaussian, so (4.8.15) also says that the expectation value of $\hat N$ is non-negative in every state this chapter writes down. Now take $\ket\psi$ to be the eigenvector $\ket\nu$ itself. The right-hand side is $\nu\norm{\ket\nu}^{2}$, the norm on the right is a positive number because $\ket\nu$ is not the zero vector, and therefore

No eigenvalue of $\hat N$ is negative. That is the bound the ladder needs, and notice what produced it. Not the form of the potential, not the behaviour of a wavefunction at infinity, not any claim about energies being positive. The positivity of the spectrum of $\hat a^{\dagger}\hat a$ is the positivity of $\avg{v,v}$, which is an axiom of the inner product and was written down in Chapter 0.5 before quantum mechanics appeared in this book at all.

4.3 · Closing the argument

Now put §4.1 and §4.2 together. Suppose $\nu$ is an eigenvalue and is not a non-negative integer. A negative $\nu$ is excluded by (4.8.16) at once, so the case to rule out is a positive $\nu$ that is not a whole number. Descending gives eigenvalues $\nu-1,\nu-2,\dots$, and the descent can only be stopped by hitting the zero vector, which by (4.8.16) read backwards happens only when the eigenvalue being lowered from is zero. A number that is not a whole number never becomes zero by having whole numbers subtracted from it, so the descent is never stopped and eventually produces a negative eigenvalue. That contradicts (4.8.16). Hence every eigenvalue of $\hat N$ is a non-negative integer, and we may finally call it $n$.

The converse direction needs one more line, since so far we know only which values are permitted, not which occur. Suppose a state $\ket 0$ exists with $\hat a\ket 0=0$. Then $\hat N\ket 0=\hat a^{\dagger}\hat a\ket 0=0$, so it has eigenvalue zero, and climbing from it produces eigenvalue $n$ at the $n$-th step provided the climb never dies. It never does: by the same move as (4.8.15), with $\hat a\hat a^{\dagger}=\hat N+1$ from (4.8.10),

So each rung is genuinely occupied, and the ladder is unbounded above and stops dead below. All of it rests on a state annihilated by $\hat a$ existing at all, which is not yet proved. Section 5 proves it by solving one first-order differential equation, and finds exactly one solution.

4.4 · The spectrum

Substituting the permitted values of $\hat N$ into (4.8.12) gives the result the chapter was built to reach:

Everything that went into it was: the canonical commutator, the definition of the adjoint, and the non-negativity of a norm. No series was summed and no boundary condition was imposed, and the only differential equation anywhere in the chapter is §5.2's, which is first order and settles nothing but whether the bottom rung exists. That is the sense in which this is a different technique and not a shorter road to the same one.

Three debts fall due here. Chapter 0.8's closing brick promised ladder operators, and the $\tfrac12\hbar\omega$ that will not go away, and it does not go away because §2.2 showed it is the commutator. Chapter 1.3 §4.4 imposed the Bohr–Sommerfeld condition on the classical phase-space area, got $E_{n}=(n+\tfrac12)\hbar\omega$, and marked the condition itself as quoted forward to this chapter, saying the result would be derived here exactly, with ladder operators and no semiclassical approximation. It has been. Section 6 closes that loop from the other end by computing the area. And Chapter 4.5 §5.2, having produced the same eigenvalues from the Hermite side, said that this chapter would produce them a second way and get the same number.

The last of the three is the reason for doing a calculation twice, and it takes a paragraph. The two derivations share no step. Chapter 4.5 defined a family of polynomials by a generating function, differentiated it twice to get recurrences, assembled a differential equation and read the eigenvalue off the coefficient. This chapter wrote one commutator and used the fact that a length is not negative. They have the operator in common and nothing else, and they agree. An answer reached twice by disjoint routes carries more weight than an answer reached once by either, and more still when one of the routes generalises to field theory and the other does not.

4.5 · The rungs, normalised

Before going further we need to know not only that $\hat a^{\dagger}$ moves up a rung but by how much it stretches the vector, because §§5 to 7 all compute with these. Equation (4.8.17) already has the answer for the upward move, and running the same line downward with (4.8.15) gives $\norm{\hat a\ket n}^{2}=n$ when $\ket n$ is normalised. So $\hat a^{\dagger}\ket n$ has length $\sqrt{n+1}$ and $\hat a\ket n$ has length $\sqrt n$. Each is a multiple of the neighbouring normalised state, and only one of the two multiples is free. Define $\ket{n+1}$ to be $\hat a^{\dagger}\ket n/\sqrt{n+1}$, which fixes that phase to $+1$ at every rung. The downward relation then follows rather than being chosen a second time, since $\hat a\ket{n+1}=\hat a\hat a^{\dagger}\ket n/\sqrt{n+1}=(\hat N+1)\ket n/\sqrt{n+1}=\sqrt{n+1}\ket n$, with the same sign:

The second relation at $n=0$ reads $\hat a\ket0=0$, which is the terminating condition of §4.3 written in the new notation. Iterating the first from the bottom expresses every state in terms of the lowest one, with the factorials coming from the product of the square roots:

One thing this does not yet establish is that each rung holds exactly one state. The algebra shows that a state at rung $n$ descends to a state at rung $0$, so the number of independent states at rung $n$ is at most the number at rung $0$, which is the dimension of the space of solutions of $\hat a\ket0=0$. That dimension is shown to be one in §5.2 below, and non-degeneracy follows.

a natural place to stop  ·  the spectrum is derived and the algebra is finished; what follows turns the same two operators on the wavefunctions, the classical orbit, and a state that moves

5 · The wavefunctions, from $\hat a\ket0=0$

Here is where this section is going. We have a spectrum but no functions, and one thing still owed from §4.3: that a state killed by $\hat a$ exists and is unique. Both are settled by the same short calculation, and the calculation is the payoff of the whole method. A second-order eigenvalue problem with infinitely many solutions to find has become one first-order equation, solved once, plus an algebra that generates the rest. By the end of the section the states will be identified with Chapter 4.5's Hermite functions, and their parities will have been read off without looking at any of them.

5.1 · One first-order equation

To turn $\hat a\ket0=0$ into something solvable we need $\hat a$ as a differential operator, which means going to the position representation. Chapter 4.6 §5 established that there $\hat p$ acts as $-\ii\hbar\,\dd/\dd x$ and $\hat x$ as multiplication by $x$. Substituting both into (4.8.7) and writing everything in the dimensionless $\xi=x/x_{0}$ turns the operator into

The relative sign is the one place to be careful, and it comes from the $\ii$ in (4.8.7) meeting the $\ii$ in $\hat p$. Checking (4.8.10) against these two expressions directly costs one line: applying $(\xi+\dd/\dd\xi)(\xi-\dd/\dd\xi)$ and its reverse to a test function and subtracting leaves $2$, and the halves in front make it $1$.

5.2 · Solving it, and why the solution is unique

Now write out $\hat a\psi_{0}=0$. The $1/\sqrt2$ multiplies zero and can be dropped, leaving

This is a first-order linear equation with separable variables, which Chapter 0.8 §2.1 solved in general: divide by $\psi_{0}$, integrate both sides, exponentiate. Doing that gives $\ln\psi_{0}=-\xi^{2}/2+\text{const}$, and fixing the constant with Chapter 0.2 §4's Gaussian integral,

The $x_{0}^{-1/2}$ is the whole difference between the two forms, and every wavefunction after this one is written in the second of them. The left-hand form is normalised against $\dd x$ and therefore carries the dimensions of an inverse square root of length. The right-hand form pulls that dimension out into $x_{0}^{-1/2}$ and leaves a dimensionless function of $\xi$, normalised against $\dd\xi$. From here on the states are the dimensionless factors, and getting back to a function of $x$ means restoring the $x_{0}^{-1/2}$ and reading $\xi$ as $x/x_{0}$.

Two things follow at once. First, the ground state is a Gaussian, which is going to matter twice more in this chapter, at §6.1 and again at §7. Second, and this is what §4.5 was waiting for, the solution is unique up to a constant multiple. Chapter 0.8 §3.1 showed that the solution space of a homogeneous linear differential equation has dimension equal to its order, and (4.8.22) is of order one. So $\hat a$ kills a one-dimensional space of functions, there is exactly one normalised ground state, and therefore exactly one state at every rung. The oscillator's spectrum is non-degenerate, and the reason is that a first-order equation has a one-parameter family of solutions.

Chapter 0.8 opened by promising that its central sentence, that a linear differential equation is an eigenvalue problem, would appear three times in different costumes, and named this chapter as one of them. Here is the costume. The eigenvalue problem that does the work is not the second-order one that the Hamiltonian looks like. It is (4.8.22), first order, with the eigenvalue zero, and an algebra supplies everything else.

The effort saved is easy to measure. The eigenvalue problem for (4.8.1) is a second-order equation, one for each of infinitely many eigenvalues, with normalisability deciding which of them are allowed. What replaced it was (4.8.22), solved once in three lines, plus (4.8.20).

5.3 · Climbing

With the ground state in hand, every other state follows from (4.8.20). In the position representation that reads

Doing the first two by hand shows the pattern. Applying the operator once, the derivative of the Gaussian brings down a factor $-\xi$, which adds to the $\xi$ already there: $\psi_{1}=\sqrt2\,\xi\,\pi^{-1/4}\ee^{-\xi^{2}/2}$. Applying it again and dividing by $\sqrt2$ gives $\psi_{2}=(2\xi^{2}-1)\pi^{-1/4}\ee^{-\xi^{2}/2}/\sqrt2$. In every case a polynomial multiplies the same Gaussian, and the polynomial gains one degree at each step, because the operator either multiplies by $\xi$ or differentiates.

5.4 · These are the Hermite functions, and that is a derived fact

The polynomials just generated look like the Hermite polynomials of Chapter 4.5, and they are, but saying so needs an argument rather than an appeal to appearance. The argument is one line and it uses a result that chapter proved about polynomials, not about eigenvalues.

Chapter 4.5 §5.2 defined $H_{n}$ by a generating function and derived, in its grind box, the identity $H_{n+1}=2\xi H_{n}-H_{n}'$. Write $u_{n}=H_{n}(\xi)\ee^{-\xi^{2}/2}$ and apply the raising operator of (4.8.21) to it. The derivative hits both factors, and the Gaussian contributes $-\xi u_{n}$, which cancels one of the two $\xi$ terms:

The raising operator is the Hermite recurrence. That is the identification, and the normalisation follows from it. Chapter 4.5 §5.1 normalised its functions as $h_{n}=u_{n}/\sqrt{2^{n}n!\sqrt\pi}$, and dividing (4.8.25) by that constant and by the $\sqrt2$ in (4.8.21) turns the coefficient into $\sqrt{2^{n+1}(n+1)!}/(\sqrt2\sqrt{2^{n}n!})=\sqrt{n+1}$, which is precisely (4.8.19). Since $h_{0}=\pi^{-1/4}\ee^{-\xi^{2}/2}$ agrees with (4.8.23), induction gives $\psi_{n}=h_{n}$ for every $n$, including the sign and the constant.

What has and has not been imported needs stating precisely. The recurrence used above is a statement about a family of polynomials. It says nothing about energies. The eigenvalues of this chapter came out of (4.8.10) and (4.8.16) and would be what they are if Chapter 4.5 had never been written. What §5.4 adds is that the functions produced by the two routes are the same functions, which is a check the calculation could have failed and did not.

5.5 · Completeness, cited and not reproved

One question remains and this chapter does not answer it, because it has already been answered. Are these all the states? The ladder produces one state per rung and the algebra says nothing about whether some other vector of the space lies outside all of them. Chapter 4.5 §5.4 proved that no non-zero vector is orthogonal to every Hermite function, its §5.5 used that to complete the spectral verification for this exact operator, and its §5.6 stated how far the verification reaches. Chapter 4.3's closing brick promised in writing that this would happen, and said what would follow from it, which is that the present chapter expands in the Hermite functions freely.

So we do, from here on, and the licence is Chapter 4.5's rather than this chapter's. Any state of the oscillator may be written $\ket\psi=\sum_{n}c_{n}\ket n$ with $\sum\abs{c_{n}}^{2}$ finite, the sum converges in the norm, and Chapter 4.5 §8.2 lists what may then safely be done with it. Section 7 uses that licence hard, and it is the only place in the chapter that needs it. This also collects a promise made a long way back: Chapter 0.5 said that applying its orthogonalisation procedure with a weight $\ee^{-x^{2}}$ on the whole line produces the Hermite polynomials, and that those are the quantum harmonic oscillator states of this chapter. They are, and (4.8.24) is how they got here without an orthogonalisation being run.

5.6 · The parities, by a third route

Chapter 4.7 §2 built parity, the operator $\hat\Pi$ that reflects a function through the origin, and showed it commutes with the Hamiltonian whenever the potential is even. The oscillator's is, so its states must each be even or odd, and Chapter 4.7 §3.3 said this chapter would find the alternation by a third route. Here it is, and the route uses no wavefunction.

Reflection reverses position and momentum, which in operator terms is $\hat\Pi\hat x\hat\Pi=-\hat x$ and $\hat\Pi\hat p\hat\Pi=-\hat p$, using $\hat\Pi^{2}=\hat I$. Both appear linearly in (4.8.8), so the raising operator reverses too:

the second because (4.8.23) is a Gaussian in $x^{2}$ and is therefore unchanged by $x\to-x$. Now apply $\hat\Pi$ to (4.8.20) and slide it through the $n$ copies of $\hat a^{\dagger}$, inserting $\hat\Pi^{2}=\hat I$ between each pair. Each passage costs a minus sign and the last $\hat\Pi$ lands on $\ket0$ and does nothing, so

The parities alternate up the ladder starting even, exactly as they did in the infinite well and in the finite well. Three routes to the same fact now exist: Chapter 4.7 §2.4 got it from the non-degeneracy of bound states in one dimension, its §4.3 got it from which branch of a transcendental condition each root lies on, and this got it from moving one operator past another. Chapter 4.7 §2.5 listed the places parity gets spent, and Chapters 4.16 and 4.17 will spend it on states of this chapter's ladder in particular.

One corollary is free. The degree of $H_{n}$ is exactly $n$, which Chapter 4.5 §5.4 established from the recurrence, so $\psi_{n}$ has at most $n$ zeros. Combined with (4.8.27), those zeros come in pairs at $\pm\xi$, apart from a possible one at the origin, which every odd state has. The numerically computed states in §6.5's figure carry exactly $n$ of them. The infinite well gave the same count, but the two chapters start their counters in different places, so the agreement needs stating in a form neither convention owns. In both problems, the state $k$ rungs above the lowest has exactly $k$ interior zeros. Chapter 4.7 numbers that state $n=k+1$ and this chapter numbers it $n=k$.

6 · The phase-space area, collected

Here is where this section is going. The classical oscillator of Chapter 0.8 traced an ellipse in phase space, and the area of that ellipse was computed there and left with an explicit promise attached to it. This section pays the promise. First we measure how large a quantum state is in position and in momentum, which turns out to place it exactly on the classical ellipse of the same energy. Then we compute the enclosed area and find that it is $(n+\half)h$, which is what Chapter 1.3 guessed semiclassically and what this chapter has now derived without guessing. The section ends with a numerical check of the entire spectrum that shares no step with any of the above.

6.1 · How big is a state

We want the spread of position and of momentum in the state $\ket n$, and (4.8.9) makes both easy, because it turns a question about $\hat x$ into a question about which rungs the operators connect. Start with the means. Since $\hat a$ lowers a rung and $\hat a^{\dagger}$ raises one, both send $\ket n$ to a state orthogonal to $\ket n$, so $\avg{n|\hat x|n}$ and $\avg{n|\hat p|n}$ are both zero, at every rung. No energy eigenstate of an oscillator is anywhere in particular, and none of them is moving. Section 7 exists to fix that.

For the squares, expand $\hat x^{2}$ from (4.8.9). Four products appear. The terms $\hat a^{2}$ and $(\hat a^{\dagger})^{2}$ move two rungs and contribute nothing to a diagonal expectation value, and the two surviving terms are $\hat a\hat a^{\dagger}=\hat N+1$ and $\hat a^{\dagger}\hat a=\hat N$, whose sum in the state $\ket n$ is $2n+1$:

The second comes out of the same expansion. Squaring the momentum in (4.8.9) puts an overall minus sign in front, from the $\ii^{2}$, and also flips the sign of the two terms that survive on the diagonal, so the two changes cancel and the answer differs from the first only in carrying $p_{0}^{2}$ rather than $x_{0}^{2}$. Two consequences are immediate. Since the means vanish, the spreads of Chapter 4.2 §5.3 are the square roots of these, and their product is

At the bottom rung that is $\hbar/2$, and that number has already been met. Chapter 0.9 §6's bandwidth theorem bounds the product of a function's width and its transform's width from below, and Chapter 4.6 §10.2 turned that into $\Delta x\,\Delta p\ge\hbar/2$ using the single substitution $p=\hbar k$. So the ground state sits exactly on a bound proved before quantum mechanics entered the argument. Chapter 0.9 §6.5 went further and showed that the only functions attaining it are Gaussians, and (4.8.23) is a Gaussian, so the two statements agree and neither was arranged to. The general inequality for an arbitrary pair of observables belongs to Chapter 4.9 §2 and is not needed here. What this section has is one instance, computed exactly, before the theorem arrives, which is the order this book prefers wherever it can get it.

The other consequence is a check on the arithmetic that costs nothing. The mean kinetic energy is $\avg{\hat p^{2}}/2m=\tfrac14(2n+1)\hbar\omega$ and the mean potential energy is $\half m\omega^{2}\avg{\hat x^{2}}$, which is the same number. So the energy divides equally between the two, at every rung, and the sum is $E_{n}$ as it must be.

6.2 · The ellipse, rationed

Now bring in the classical picture, because the numbers in (4.8.28) are about to land on it exactly. Chapter 0.8 §4.4 plotted the classical oscillator's state as a point $(x,p)$ and found that energy conservation confines it to an ellipse, $p^{2}/2mE+x^{2}/(2E/m\omega^{2})=1$. Substitute the two quantum averages into the left-hand side, at energy $E=E_{n}$. Each term contributes exactly one half, and

The root-mean-square point of the quantum state sits on the classical orbit of the same energy. That is not an approximation and it does not require $n$ to be large. It holds at $n=0$, where the state has no classical counterpart at all.

Chapter 0.8 §4.4 then computed the area the ellipse encloses, getting $\mathcal A=2\pi E/\omega$, and Chapter 1.3 §4.4 wrote the same quantity as the loop integral $\oint p\,\dd q$ and made it the central object of its phase-space section. Verified symbolically: integrating $p=\sqrt{2m(E-\half m\omega^{2}q^{2})}$ around the closed orbit returns $2\pi E/\omega$ exactly. Now put the derived spectrum (4.8.18) into it, and use $h=2\pi\hbar$:

Three promises are paid by that line. Chapter 0.8 §4.4 said that the classical ellipse is not replaced by quantum mechanics but rationed, and its closing brick called the area the area that Chapter 4.8 will quantise. Chapter 1.3 §4.4 marked the Bohr–Sommerfeld condition as quoted forward to here. And Chapter 1.3 said of its three observations that this chapter would confirm them by an exact operator calculation that uses none of this reasoning, which is a strong claim. Sections 2 to 4 used a commutator, an adjoint and the non-negativity of a norm. No orbit appeared, no adiabatic invariant, no correspondence argument and no appeal to large quantum numbers. The claim holds.

Equation (4.8.31) reads two ways. Successive allowed orbits enclose areas differing by exactly $h$, so the ring between one orbit and the next has area equal to Planck's constant, always, everywhere on the ladder. And the innermost orbit encloses $h/2$ rather than nothing, which is the zero-point energy of §2.2 drawn as a picture: the state cannot be a point in phase space, and the smallest region it can occupy is half a unit of $h$.

6.3 · What is this chapter's, and what is Chapter 4.10's

A boundary needs drawing here, so that you know exactly what has been established and what has not. Chapter 1.3's Bohr–Sommerfeld condition is a general statement about any bound one-dimensional motion, and it was a guess when it was made, supported by the argument that an adiabatic invariant is the right thing to quantise. This chapter has derived one case of it, exactly, and has said nothing about any other case.

The general statement is Chapter 4.10 §6, where it comes out of a genuine approximation scheme rather than a guess, and where it is scored: for potentials that are not quadratic the condition is approximate, the error is largest at the bottom of the ladder, and that chapter prints the numbers. So the oscillator is the case where the semiclassical answer is exact, and the reason it is exact is that the answer came from an algebra with no small parameter in it. Chapter 4.10 also supplies a count that Chapter 4.1 needed and could not get. That chapter's classical partition function had to divide phase space into cells before it could count states, and classical mechanics offered no size for a cell, so the sum was defined only up to a constant. The count that closes it is that an orbit enclosing area $\mathcal A$ holds about $\mathcal A/h$ states, which is (4.8.31) read as a density rather than as a level, and 4.10 is where it is proved.

6.4 · The numerical confirmation

Everything above came out of one commutation relation, so the whole of it can be held against a computation that has never heard of that relation. The check used here is a direct diagonalisation of (4.8.1) in position space, with no ladder operator, no Hermite polynomial and no generating function anywhere in it.

The construction is short to describe, and the one choice in it needs a reason. Work in units $\hbar=m=\omega=1$ and put down a uniform grid of $60$ points at spacing $0.3$, spanning $\abs x\le9$. What has to be represented on that grid is the kinetic energy $-\half\,\dvn{2}{}{x}$. An ordinary finite difference would represent it only to some power of the spacing, and the comparison below would then be measuring the difference scheme rather than the spectrum. So use instead the matrix that is exact on every function whose Fourier content fits between the grid's own limits. Its entries are $\pi^{2}/3$ on the diagonal and $2(-1)^{j-k}/(j-k)^{2}$ off it, all divided by twice the squared spacing. The $-\half$ is already inside them, so what those entries define is the kinetic-energy operator and not the second derivative, which is why the next step adds the potential rather than subtracting it. Add $\half x^{2}$ down the diagonal and diagonalise the resulting $60\times60$ symmetric matrix. The whole computation is a discretisation of the differential operator and a rotation to its eigenbasis.

The lowest sixteen eigenvalues come out as $0.5,1.5,2.5,\dots,15.5$, with the largest departure from $(n+\half)$ anywhere in that range equal to $1.2\times10^{-13}$ and the largest departure of any gap from $1$ equal to $6.2\times10^{-14}$. Both are at the level of double-precision rounding for a matrix of this size, so the levels are equally spaced as far as the arithmetic can see. Taking the same eigenvectors and computing $\sum_{j}c_{j}^{2}x_{j}^{2}$, which is $\avg{\hat x^{2}}$ read straight off the grid, reproduces (4.8.28) to a worst relative departure of $1.5\times10^{-14}$ over the same sixteen states, and the corresponding $\Delta x\,\Delta p$ reproduces (4.8.29) to twelve figures at every one of them.

Two remarks about what that does and does not test. The accuracy is not an accident of a fine grid: it comes from the eigenfunctions decaying faster than any exponential, so a grid that comfortably contains them and comfortably resolves them commits an error that falls off faster than any power of the spacing. The grid has to do both, and it fails at each end for its own reason. Shrinking the span to $\abs x\le7.5$ at the same $60$ points degrades the sixteenth level to $4\times10^{-7}$, because the highest states no longer fit inside it. Stretching it to $\abs x\le14$ degrades the same level to $1\times10^{-3}$, because the spacing is then too coarse to follow their oscillations. Between those two the window is wide, and inside it the answer is right to the level of double-precision rounding quoted above. The test is also genuinely independent, since a grid Laplacian shares no step with §§2 to 4. What it does not test is completeness, which is Chapter 4.5's and cannot be settled by any finite matrix.

6.5 · The ladder, operable

The figure below is the ladder with a button on it. Press $\hat a^{\dagger}$ and the state climbs a rung, press $\hat a$ and it falls, and at the bottom pressing $\hat a$ returns nothing at all, which is (4.8.19) at $n=0$ and is the whole reason the spectrum has a floor. The levels, the wavefunctions and the expectation values are all taken from §6.4's diagonalisation rather than from any formula in this chapter, so the readouts are testing the algebra rather than displaying it.

raise  â^† lower  â ground state

n = 0 E_n = 0.5000000000000 hbar omega E_n - (n + 1/2) = 8.33e-15 largest |gap - hbar omega| over all 16 levels = 6.22e-14 -- diagonalised on a grid, with no ladder operator in it

<x^2> from the eigenvector = 0.500000000000 x0^2 (n + 1/2) = 0.5 relative difference = 8.88e-15 Dx Dp = 0.500000000000 hbar enclosed area = 0.500000000000 h

the ground state, which is the state a annihilates

a natural place to stop  ·  the energy eigenstates are finished; what follows builds a state that is not one of them, and it is the state that behaves classically

7 · Coherent states

Here is where this section is going. Six sections have produced energy eigenstates, and §6.1 noted a strange fact about all of them: in every one, the mean position and the mean momentum are zero, and they stay zero forever, because a stationary state's expectation values do not move. Nothing built so far swings. This section builds the state that does, finds that its width never changes while it swings, and observes that it is not an energy eigenstate at all.

7.1 · What to ask for

The wanted state should behave as much like a classical oscillator as quantum mechanics permits, and the natural way to look for it is to ask what the classical variables become. Put Chapter 0.8's solution $x=A\cos(\omega t+\varphi)$ into the combination $x+\ii p/m\omega$, using $p=m\dot x$, and the two terms assemble into $A\ee^{-\ii(\omega t+\varphi)}$. Its modulus is the amplitude and its argument turns at the oscillator's own rate, which is that chapter's §4.2 complex amplitude set rotating and its §4.4 circle written as one number. That combination is precisely what (4.8.7) was built out of, so ask for a state on which $\hat a$ acts as a number:

The warning in §2.4 is what makes that request legitimate. Because $\hat a$ is not self-adjoint, nothing requires $\alpha$ to be real, and none of Chapter 4.2's theorems about the eigenvalues of an observable applies. There is no measurement whose outcome is $\alpha$. What there is, if such states exist, is a family labelled by a complex number, and the label should turn out to be the classical amplitude.

7.2 · Building it

To find the states, expand in the basis we have. Chapter 4.5 §5.4's completeness is what permits the expansion, and §5.5 above collected it, so write $\ket\alpha=\sum_{n\ge0}c_{n}\ket n$ and apply (4.8.19) term by term. Lowering shifts every index down one, so

Now compare the coefficient of $\ket n$ on the two sides, which is legitimate because the $\ket n$ are orthonormal. On the left the term with $\ket n$ comes from $n+1$, so $c_{n+1}\sqrt{n+1}=\alpha c_{n}$. That recurrence unwinds immediately, giving $c_{n}=\alpha^{n}c_{0}/\sqrt{n!}$, and the constant $c_{0}$ is fixed by demanding $\sum\abs{c_{n}}^{2}=1$, where the sum is the exponential series. So the state exists, for every complex $\alpha$, and it is

Those coefficients also settle the domain question of §2.4 for this state, which is not a finite combination and so is not covered by what that box said. Since $\abs{c_{n}}^{2}=\ee^{-\abs\alpha^{2}}\abs\alpha^{2n}/n!$ falls off faster than any power of $n$, the sum $\sum_{n}n^{k}\abs{c_{n}}^{2}$ converges for every $k$, so $\ket\alpha$ lies in the domain of every polynomial in $\hat a$ and $\hat a^{\dagger}$. That is what licenses moving $\hat a$ across the inner product in §7.3 and §7.5, and again in Problem 3(a).

7.3 · Its position, its momentum, and its width

What we want to know first is where this state is and how wide it is, and both come from (4.8.9) with (4.8.32) used on the right and its adjoint on the left. Since $\hat a\ket\alpha=\alpha\ket\alpha$ and $\bra\alpha\hat a^{\dagger}=\bar\alpha\bra\alpha$, every expectation value reduces to arithmetic in $\alpha$:

So the label is the position and momentum of the state, in the natural units, and the state can be put anywhere. For the widths, use the same trick on $\hat x^{2}$, being careful with the one term that is not already in normal order: write $\hat a\hat a^{\dagger}$ as $\hat a^{\dagger}\hat a+1$ before evaluating, which is (4.8.10) earning its keep for the last time. The cross terms combine into $\avg{\hat x}^{2}$ exactly, and the $+1$ survives on its own:

What is missing from those expressions is the point: the widths do not depend on $\alpha$. Every state of this family, however far from the origin it is placed and however fast it is moving, has exactly the width of the ground state, and saturates Chapter 0.9 §6's bound. That is consistent with §6.1, where the ground state saturated it too, and indeed $\alpha=0$ in (4.8.34) gives $\ket0$ back. The coherent states are the ground state, moved.

7.4 · The state that does not spread

Now let it evolve, which is where the point of the whole section is. Chapter 4.6 §9 established that a stationary state acquires the phase $\ee^{-\ii E_{n}t/\hbar}$ and nothing else, so each term of (4.8.34) picks up $\ee^{-\ii(n+\half)\omega t}$. Pull the half out front, where it is a global phase and affects nothing measurable, and what is left multiplies $\alpha^{n}$ by $\ee^{-\ii n\omega t}$, which is the same as replacing $\alpha$ by $\alpha\ee^{-\ii\omega t}$ throughout:

A coherent state evolves into a coherent state, with its label rotating at the classical frequency. Everything follows from that one line. Since $\abs\alpha$ never changes, the widths in (4.8.36) never change either, and the packet does not spread. And feeding the rotating label into (4.8.35), with $\alpha=\abs\alpha\ee^{\ii\theta}$,

which is Chapter 0.8's classical solution exactly, amplitude and phase and all. The state is a rigid Gaussian of fixed width sliding back and forth along the classical trajectory.

This settles a debt from the chapter before last. Chapter 4.6 §10 built a free Gaussian packet and watched it spread, and its closing brick said that this chapter would do the same for the oscillator and find the state whose width does not change. That state is (4.8.34), its width is (4.8.36), and in the units that chapter's figure uses, where $\hbar=m=\omega=1$, the number is $1/\sqrt2$. That figure's oscillator panel was run at exactly that width for exactly this reason.

Equation (4.8.38) is easy to over-read, and two boundaries keep it in place. Chapter 4.9 §5 will show that the mean position of any state follows the classical trajectory exactly whenever the potential is at most quadratic, so the motion of the centre is not special to coherent states. What is special is the width. And no state of any system is genuinely classical: (4.8.36) is a fixed non-zero spread, and §7.5 says what else is spread.

7.5 · Not an energy eigenstate, and that is the point

After six sections of eigenstates, (4.8.34) is not one of them, and that needs saying plainly. It is a superposition of all of them, with weights that never vanish, and its energy is not definite. The mean rung follows from $\avg{\hat N}=\avg{\alpha|\hat a^{\dagger}\hat a|\alpha}=\abs\alpha^{2}$, and the spread from the same normal-ordering step used in §7.3:

The probability of finding rung $n$ is $\abs{c_{n}}^{2}=\ee^{-\abs\alpha^{2}}\abs\alpha^{2n}/n!$, which is a Poisson distribution with mean $\abs\alpha^{2}$, and (4.8.39) is its standard deviation written out. That is the reason a coherent state looks classical when it is large. The fractional uncertainty in the energy falls like $1/\abs\alpha$, so a state carrying a million quanta has its energy defined to a part in a thousand, and a state carrying one does not have a defined energy in any useful sense.

7.6 · Where this goes

Chapter 0.8 §7 closed with a sentence this chapter can now honour. A field in a box is a collection of independent modes, each one a harmonic oscillator with its own frequency, so quantising a field means doing the calculation of §§2 to 4 once and attaching a copy of it to every mode. The integer labelling a rung then counts how many quanta of $\hbar\omega$ that mode holds, and those quanta are what we call particles. A photon is one rung of one mode of the electromagnetic field, and Chapter 5.3 is where that sentence becomes a construction. Chapter 7.4 runs the same machine on the vibrational modes of a string, where the rungs are the particle spectrum. Chapter 0.3's closing list named all three destinations in order, and this chapter is the first of them.

Coherent states are the piece that travels furthest in one respect. A field mode in a coherent state has a definite amplitude and phase and an indefinite number of quanta, which is what a classical electromagnetic wave is, and it is why the classical limit of a quantum field is a coherent state rather than a state of many photons with definite number. The half rung per mode travels too, and less comfortably: every mode of every field carries $\half\hbar\omega$ whether anything is in it or not, and adding that up over infinitely many modes gives infinity. Chapter 5.3 has to say what to do about it before it can say anything else.

8 · Worked examples

9 · Your turn