# Position operator and Momentum operator in the Energy basis

In studying Shankar quantum mechanics p.208 on expressing matrix elements of position operator and momentum operator in terms of the energy basis of the harmonic oscillator Hamiltonian $$H=\frac{P^{2}}{2m} + \frac{1}{2} m\omega^{2}X^{2}$$ , a question came up to my mind.

Cleary, $$H$$ eigenbasis $$|n\rangle$$ can incorporate in its eigenspace only the states that are some superposition of these eigenvectors, that is, physically realizable states residing in the Hilbert space to the physical problem at hand.

But evidently, $$X$$ basis and $$P$$ basis have capabilities of spanning vector spaces (as their eigenspaces) with dimensions equal to $$R$$ whereas $$H$$ eigenbasis forms a vector space which has dimension equal to $$N$$. That being said, I think there can be possibilities in which upon applying either position operator $$X$$ or momentum operator $$P$$ to one of the energy eigenstate $$|n\rangle$$, these operators might yield as a resulting state the one that does not belong to $$H$$ eigenspace. And this at the same time means that $$X$$ and $$P$$ operators may not be expressed as some infinite dimensional matrix with its matrix elements written in terms of $$H$$ basis of the oscillator Hamiltonian.

I think my question here is pretty reasonable, but what am I missing or have mistaken here?

• Is your problem that Energy is discrete put X and P are continuous? Bc they both span vector spaces, Im not exactly sure what you mean by R and N. The whole point of the Harmonic Oscillator is to show that the Energy Eigenstates are discrete, and each one has a position/momentum distribution associated with it; because of that our particle MUST obey certain statistical rules given its Energy eigenstates – xXx_69_SWAG_69_xXx Jan 26 at 16:27
• Indeed, N is countable and R is uncountable, but, then again, X eigenstates are not normalizable and in the Hilbert space, even though they serve to define basis states in the Hilbert space. The real basis states are the normalizable Hermite functions, $\langle x |n\rangle$, which are a countable set! – Cosmas Zachos Jan 26 at 16:36
• – Cosmas Zachos Jan 26 at 16:55
• This question really has nothing to do with quantum mechanics. Ask yourself instead this: how is it possible that a continuous function $f: [0, 1] \rightarrow \mathbb{R}$ defined over the interval $[0, 1]$ can be expressed as a discrete sum Fourier series $\sum_{k=-\infty}^\infty c_k \exp(i 2 \pi k x)$? – DanielSank Jan 26 at 17:39
• – Cosmas Zachos Jan 26 at 21:31

Right you are: the set $$\{ |x\rangle \}$$ is uncountable and the set $$\{ |n\rangle \}$$ is countable. But, when discussing the oscillator, we are not moving around all xs. We are really moving in the Hilbert space of normalizable states, the (real) Hermite polynomials, $$\psi_n(x)=\langle x| n\rangle$$, a complete orthonormal countable set, s.t. $$|n\rangle= \int\!\! dx ~|x\rangle \langle x|n\rangle = \int\!\! dx ~~\psi_n(x) |x\rangle$$ square integrable, normalized, etc. So we "pretend" $$|x\rangle = \sum_{n=0}^{\infty} |n\rangle \langle n|x\rangle = \sum_{n=0}^{\infty} \psi_n(x) |n\rangle ,$$ as we are not really considering the entire domain of $$\hat X, \hat P$$. We are considering the countable part that has eigenvalues $$n+1/2$$ for the hamiltonian $$(\hat {X}^2+ \hat P^2)/2$$ (where we've absorbed the obnoxious constants $$\hbar, m, \omega$$ into our normalizations).

We are really moving in a Hilbert space of infinite discrete matrices, as you saw, and your text sensibly doesn't dwell on this projection by the complete Hermite functions. There is an entire continent of "Rigged Hilbert spaces", but I haven't appreciated if you really want to go there. Most practical texts don't.

NB on comments (geeky)

The above projection on energy (number) eigenstates is consistent, to the extent the action of $$\hat X= (a+a^\dagger)/\sqrt{2}, \qquad \hat P= (a^\dagger - a) /\sqrt{2}$$ on $$|n\rangle$$ raises or lowers n by 1, and so maintains the projection on integer ns, as one may also confirm from the recursion relations of Hermite functions, linked above.

So, the dynamics never slips into the unphysical superselection sectors projected out, e.g., with m = integer plus a noninteger constant, such as 1/2. Freak states such as $$|1/2\rangle \equiv \sqrt{ \frac{a^\dagger}{(1/2)!}}|0\rangle= \sqrt{ \frac{2a^\dagger}{ \sqrt{\pi}}}|0\rangle$$ orthogonal to the above integer n ones are permanently and safely excluded from consideration.

• I think the correspondence between smooth functions and discrete bases is more general that even this answer suggests. Just think of the Fourier series: $f(x) = \sum_{n=-\infty}^\infty c_n \exp(i 2 \pi n x)$. – DanielSank Jan 26 at 18:17
• @DanielSank Your point rules, of course, but your compact interval simplifies the issue. The compactness enforced by the oscillator potential is "diffused" all over, and first evident in the quantized spectrum. Not being Cantor, I can't cook up counterexamples of functions not resolvable in Hermite functions. – Cosmas Zachos Jan 26 at 18:41
• Ahhh, good point. – DanielSank Jan 26 at 19:25
• Thank you so much for answering my question! I tried on the text you linked me about rigged Hilbert space but ended up finding myself helpless.... But I see your point that we are not considering the entire domain of X and P and we are moving only in the Hilbert space of Hermite polynomials. Still one doubtful point remains though. How can we tell that the set spanned by Hermite polynomials is closed under operation of X and P? – 류민석 Jan 27 at 7:20
• Appreciate it, sir! – 류민석 Jan 27 at 11:14