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I'm self-studying QM and have a basic question on quantum harmonic oscillator. The Hamilton is certainly quantized under this model, that is $E_n=(n+1/2)\hbar \omega$, for $n=0,1,2,...$. But is linear momentum quantized under this model? The textbook by Griffith told me momentum operator has continuous spectra, so it shouldn't be quantized.

But $L^2(\mathbb{R})$ (square-integrable functions) is separable, and the weighted Hermite polynomials (which are the eigenstates of the quantum harmonic oscillator) constitute a countable basis for it. The momentum operator can be represented as a "countably infinite" dimensional matrix (I know I'm making up words here). So how can this matrix with a "countably infinite" dimension has continuous spectra?

Update: thank you all for the wonderful answers! I just would like to follow up with a question to make sure I'm on the right track. I'm quoting a lecture note . It is easy to check that the ladder operators can be represented by the following matrices under the stationary states of Hamiltonian: \begin{equation*} a=\begin{pmatrix} 0 & \sqrt{1} & 0 & \cdots \\ 0 & 0 & \sqrt{2} &\cdots \\ 0 & 0 & 0 &\cdots \\ \vdots & \vdots & \vdots &\vdots \end{pmatrix} \end{equation*} \begin{equation*} a^{\dagger}=\begin{pmatrix} 0 & 0 & 0 & \cdots \\ \sqrt{1} & 0 & 0 &\cdots \\ 0 & \sqrt{2} & 0 &\cdots \\ \vdots & \vdots & \vdots &\vdots \end{pmatrix} \end{equation*} Certainly, $aa^{\dagger}$ and $a^{\dagger}a$ are diagonal.

So, my question is do $a$ and $a^{\dagger}$ have continuous spectra despite their simple form? (I guess so because $\hat{x}$ and $\hat{p}$ are linear combination of $a$ and $a^{\dagger}$. So if $\hat{x}$ and $\hat{p}$ have continuous spectral, so do $a$ and $a^\dagger$.)

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    $\begingroup$ Related: Hilbert space of harmonic oscillator: Countable vs uncountable? and links therein. $\endgroup$
    – Qmechanic
    Commented Mar 24, 2023 at 8:09
  • $\begingroup$ In what sense do the existent answers do not address your edit? $\endgroup$ Commented Mar 24, 2023 at 18:51
  • $\begingroup$ Thanks! I guess it is addressed. I simply need a confirmation that the ladder operators don't have discrete spectral as p and x to make sure my understanding is correct. $\endgroup$
    – Victor
    Commented Mar 24, 2023 at 19:05
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    $\begingroup$ You got the confirmation: the "discrete" infinite-dimensional matrix you get by subtracting your two such matrices, $\hat p=i(a^\dagger -a)/\sqrt 2$, has continuous eigenvalues p for the eigenvectors provided. Do the calculation! $\endgroup$ Commented Mar 24, 2023 at 20:02
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    $\begingroup$ Your first sentence "Recall a has a well-known continuous spectrum on coherent states" precisely answered my question. Thanks! I just lack knowledge on basic functional analysis. $\endgroup$
    – Victor
    Commented Mar 28, 2023 at 5:54

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Recall a has a well-known continuous spectrum on coherent states. The disquisitions on the rigged Hilbert space representing the continuous spectrum of $\hat p=i(a^\dagger -a)/\sqrt 2$ linked by @Qmechanic address your question more than completely.

I would just like to illustrate the standard bridge between Fock space (the discrete Hilbert space of oscillator eigenstates $|n\rangle$) and the space of (unnormalizable) momentum eigenstates $|p\rangle$ you are dealing with, $$ |p\rangle \propto e^{ip\sqrt{2} ~a^\dagger+(a^\dagger)^2/2} |0\rangle .$$ It connects the discrete state Fock space to the continuous space of momentum eigenstates. The momentum and energy operators, of course, fail to commute.

By use of the BCH identity (equivalently, the seat-of-the-pants commutator correspondence $a^\dagger \to z$, and $a\to \partial_z$), you may see that $$ \hat p | p\rangle= {i(a^\dagger - a)\over \sqrt 2} | p\rangle= p | p\rangle, $$ for an arbitrary real, continuous p, since $${i(a^\dagger - a)\over \sqrt 2} e^{ip\sqrt{2} ~a^\dagger+(a^\dagger)^2/2} |0\rangle = pe^{ip\sqrt{2} ~a^\dagger+(a^\dagger)^2/2} |0\rangle .$$ This is a standard exercise you must do, which might help you see the correspondence in concrete terms.

To fully understand this bridge, you must visit the Segal-Bargmann transform.

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Quantized/quantization may mean many things - see, e.g., How does quantization arise in quantum mechanics?, Name of concept: Replace classical variables by quantum operators. My guess from the context of the question is that the OP asks whether the momentum has discrete values in in quantum harmonic oscillator. If we are talking about the oscillator eigenstates, then the answer is no - momentum does not commute with the oscillator Hamiltonian and does not have definite values.

A common (but artificial) situation where momentum is quantized is when applying periodic boundary conditions.

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