Position operator and Momentum operator in the Energy basis

Question

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?

Answer 1

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.

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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.