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I have recently been learning about the quantum harmonic oscillator and how it is described in the language of ladder operators. At the moment the logic behind the number operator seems incomplete to me. As I understand it, the Hamiltonian can easily be shown to be

$$ \hat{H} = \hbar\omega(\hat{a}^\dagger\hat{a} + \frac{1}{2}), $$

and then by comparison with the well-known energy spectrum of a harmonic oscillator, this suggests that the eigenvalues of the number operator $\hat{N} = \hat{a}^\dagger\hat{a}$ give the (integer) energy level of the oscillator. But in actually proving that $\hat{N}$ gives the appropriate eigenvalues, the general approach seems to be to use the normalised ladder operator relations:

$$ \hat{a}|n\rangle = \sqrt{n}|n-1\rangle \\ \hat{a}^\dagger|n\rangle = \sqrt{n+1}|n+1\rangle. $$

However, to my knowledge these normalising prefactors themselves come from applying the number operator when attempting to normalise the result of an application of a ladder operator!

I would like to know if there is a method of acquiring these normalisations on the ladder operators without assuming the number operator, or alternatively if there is a way of showing the number operator works that doesn't rely on already knowing the normalisations.

Qmechanic
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baivazovsky
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    "normalized" is not the correct terminology here. Those factors are what they are; $\langle n-1| a |n\rangle = \sqrt n$, it cannot be any other number. To deduce this you run through the textbook derivations. – Andrew Steane May 02 '23 at 16:49
  • by comparison with the well-known energy spectrum of a harmonic oscillator What do you mean by that? A classical harmonic oscillator can have any energy. – Ghoster May 02 '23 at 17:06
  • @AndrewSteane I see, thank you for the correction. The only derivation of those factors of which I am aware uses the fact that the eigenvalues of the number operator give the energy level as an integer - could you point me in the direction of an alternative? – baivazovsky May 02 '23 at 19:08
  • @Ghoster I was referring to the spectrum for a quantum harmonic oscillator, which my notes had previously derived using ODE methods (Hermite equation etc). – baivazovsky May 02 '23 at 19:09
  • You can derive that energy levels form a ladder with spacing $\hbar \omega$ and ground state $\hbar\omega/2$ without invoking this $\sqrt{n}$ factor. Thus you deduce eigenvalues of number operator. From that you deduce the $\sqrt{n}$ (using normalization of the state vector to do so.) – Andrew Steane May 02 '23 at 20:19

2 Answers2

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You've got the logic of the derivation wrong, I think. I recommend re-reading it, but here's an outline of the process, and you'll see that it's not circular:

  1. Show that $\hat{H}$ can be factored as $\hat{a}^{\dagger}\hat{a}+\hat{I}/2$, where the raising and lowering operators are defined in terms of $\hat{x}$ and $\hat{p}$ as usual. Note that the number operator and $\hat{H}$ clearly commute, and so they share an eigenbasis.

  2. Derive a set of commutation relations using the commutation relations for $\hat{x}$ and $\hat{p}$. The results are $[\hat{a},\hat{a}^{\dagger}]=1$, $[\hat{a},\hat{a}^{\dagger}\hat{a}]=\hat{a}$, $[\hat{a}^{\dagger},\hat{a}^{\dagger}\hat{a}]=-\hat{a}^{\dagger}$, etc.

  3. Using the commutation relations directly, show that $\hat{a}|n\rangle$ is an eigenvector of $\hat{a}^{\dagger}\hat{a}$ with eigenvalue $n-1$. We don't know yet that $n$ is an integer, and (crucially, to show that we are not making a circular argument as the OP suggests) we don't know yet what the constant of proportionality is. That is, we know that $\hat{a}|n\rangle = C_n|n-1\rangle$, but we don't know what $C_n$ is.

  4. Argue that there must be a unique ground state $|G\rangle$ that is annihilated by $\hat{a}$, and use this to show that the eigenvalue is zero, i.e., $\hat{a}^{\dagger}\hat{a}|G\rangle = 0$. From this, we can conclude that the $n$'s are non-negative integers.

  5. Assuming that by $|n\rangle$ we mean the normalized eigenstates, calculate $\langle n | \hat{a}^{\dagger}\hat{a} |n \rangle$ in two ways to determine the value of $C_n$. Similarly, we can figure out the action of $\hat{a}^{\dagger}$ on $|0\rangle$.

hft
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march
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  • In step 3, I think your $n+1$'s should be $n-1$'s. Although your step 4 is what tends to be done in textbooks, I think it's logically fallacious - why must there be such a state $|G>$? Instead, there is a way to do it by looking at $<n| a^\dagger a |n>$ (which must be nonnegative) and $<n| (a^\dagger a)^2 |n>$ (which must equal $(n-1) <n| a^\dagger a |n>$ and must also be nonnegative, so if $n < 1$ then $a|n> = |0>$). – pwf May 03 '23 at 00:17
  • ^ Correction: $<n|(a^\dagger a)^2 |n>$ should read $<n|(a^\dagger)^2a^2 |n>$ – pwf May 03 '23 at 00:26
  • @pwf. I made the fix with the $n-1$. As for the second statement, okay, that's reasonable, but that's not really the point of the post, which is to show that the argument isn't circular (since the OP seems to think that it is). – march May 03 '23 at 04:02
  • "why must there be such a state |G>?" Because if there were not such a state you could keep on lowering the energy indefinitely (since $a|G\rangle$ would have lower energy than $|G\rangle$, because $\hat H = \hat N + 1/2$, and so on) and there could be no energetically stable system. – hft May 03 '23 at 04:25
  • (And you also know you can not keep lowering the energy indefinitely since you know that the hamiltonian is also equal to $x^2/2 + p^2/2$, which is non-negative). – hft May 03 '23 at 04:31
  • @march Fair enough, your answer is good and I'm being nitpicky, but if you don't show explicitly that there must be a unique lowest rung $|G>$ such that $<G|a^\dagger a|G> = 0$ I still think you're making an assumption about $a^\dagger a$, which is what OP was looking not to do. – pwf May 04 '23 at 16:00
  • @hft Your argument about lowering the energy indefinitely does not necessarily imply the existence of a lowest rung. It amounts to, "You have to have a bottom rung because otherwise you wouldn't have a bottom rung." It also doesn't explain why that state must be unique. Rather, you are making an assumption that is motivated by a physical argument. I acknowledge that it's the argument you often see in textbooks, but it's not rigorous, and in this case it's not necessary, since a relatively simple mathematical demonstration can be given (as I outlined in my comment). – pwf May 04 '23 at 16:09
  • @hft Your argument about eigenvalues of H being non-negative, on the other hand, is stronger, and in fact it's important to show that the eigenvalues of $a^\dagger a$ are non-negative, but I still don't think that's sufficient to show that there must be a state with $<G|a^\dagger a|G> = 0$. You could, for instance, find that $<G|a^\dagger a|G>$ is not finite, i.e. that $a|G>$ is not normalizable. – pwf May 04 '23 at 16:22
  • What guarantee do we have that the $| n \rangle$ are nondegenerate? Couldn't it be that $\hat{a} | n \rangle = \sum_k C_{n ,k} |n-1 , k\rangle$ ? – Charles Hudgins Feb 28 '24 at 09:40
  • @CharlesHudgins I don't think that that conclusion can be arrived at just via the operator formalism. There are some things that cannot be shown only using the operator formalism; instead, you have to use the position-space representation and theorems of such. For example, the set of states arrived at is complete. – march Feb 28 '24 at 16:55
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First of all, I stress that the ladder operators are not defined as $$ \hat{a}|n\rangle = \sqrt{n}|n-1\rangle \\ \hat{a}^\dagger|n\rangle = \sqrt{n+1}|n+1\rangle. $$ Rhey are instead defined in terms of the position and momentum operators with suitable normalization factors in order to find the canonical commutation relations $$[a,a^\dagger]= I\:.$$

However, in order to dispel any doubts about the existence of any logical loop involving ladder operators, I point out an independent way to find the eigenvalues $n= 0,1,2,\ldots$ of the number operator $N$. This procedure does not pass through the ladder operators (and their normalization).

Consider the Hermite function (see Applications here: Hermite polynomials) $$\psi_n : \mathbb{R} \to \mathbb{C}\:,$$ where $$\psi_n(x) := (-1)^n(2^n n!\sqrt{\pi})^{-1/2}e^{x^2/2}\frac{d^n}{dx^n} e^{-x^2/2}\:,\quad n=0,1,2,\ldots\:.$$

These functions have the following nice properties.

  1. They satisfy $L^2(\mathbb{R},dx)$ orthogonality relations $$\int_{\mathbb{R}} \overline{\psi_n(x)}\psi_n(x) dx = \delta_{mn}\:.$$

  2. $\{\psi_n\}_{n \in \mathbb{N}}$ is a Hilbert basis of $L^2(\mathbb{R},dx)$

  3. They satisfy the Hermite equation which I re-arrange suitably: $$-\frac{1}{2}\frac{d^2\psi_n}{dx^2} + \frac{1}{2}x^2 \psi_n(x) - \frac{1}{2}\psi_n(x)= n\psi_n(x)\:.$$

As an immediate consequence of 3, we see that, defining $$N := -\frac{1}{2}\frac{d^2}{dx^2} + \frac{1}{2}x^2 - \frac{1}{2}\:,\tag{1}$$ for instance on the Schwartz space, this operator turns out to be symmetric and the $\psi_n$ are a Hilbert basis of eigenvectors with eigenvalues $n\in \mathbb{N}$. This is sufficient to prove (by the Nelson theorem and the uniqueness of spectral decomposition) that $N$ is (essentially) selfadjoint on that domain and its spectrum is a point spectrum which exactly coincides with $\mathbb{N}$.

It is clear that $N$ defined in (1) is the number operator when $\hbar=\omega=1$. The general case can be treated similarly with easy re-adaptations.

Valter Moretti
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