# Infinite square well eigenstates not being compatible with the momentum operator

I've been taught that, to find a given observable, one most apply the Hermitian operator pertaining to that observable. However, the function the operator the observable is being applied to must be an eigenfunction of that operator. An example of such would be something like:

$$\psi (x,t) = A e^{-i(kx-bt)}$$

This would be an eigenfunction of the momentum operator With eigenvalue $$\hbar k$$. This literally is equivalent to $$p$$, so when I'm told that an eigenfunction of an operator is a function where $$\hat p \psi = p \ \psi$$, then I whole-heartedly agree. However, if

$$\psi (x,t) = A e^{-i(\sqrt{\tau}\ x-bt)}$$

Then, it is still an eigenfunction of the momentum operator with eigenvalue $$\hbar \sqrt{\tau}$$. This is still okay with me, as the momentum must just be that value.

What confuses me, however, is that none of the solutions to Schrodinger's equation that I've learned so far even are eigenfunctions of momentum.

For instance, the solution to the infinite square well is sinusoidal, and thus has a complex eigenvalue which makes the momentum operator not produce a real observable in this case. In addition, I'm doubtful that Hermite's polynomials, solutions to a quantum harmonic oscillator, will be eigenfunctions either. Gaussian wavefunctions aren't eigenfunctions of momentum either - although I know it doesn't have a definite momentum, clearly. This seems weird to me. Why am I being taught these things if it only works in such rare cases? Can the momentum of an infinite square well eigenstate not be measured? This does not seem obvious to me. I'm maybe close to getting it. Definite momentum is only achievable with plane waves, which is why my first example worked - but if operators only seem to work on plane waves why do we use them?

• Oct 22, 2018 at 15:33

I've been taught that, to find a given observable, one most apply the Hermitian operator pertaining to that observable.

This is where the mistake lies. Making a measurement of some observable is not performed mathematically by calculating $$\hat P\psi(x)$$ In fact, you can't do anything "to find a given observable" in general. All you can look at is expectation values: $$\langle\hat P\rangle=\langle\psi|\hat P|\psi\rangle$$ In other words, you can't determine the result of a measurement of a quantum system by operating on the state vector with the corresponding operator. This is not how QM works.

You are additionally making the assumption that your state has to be an eigenstate of an operator in order to make a measurement of the corresponding observable. The general way to approach determining the probability of measuring some observable is to expand it in the eigenbasis of the corresponding operator. Then the "coefficients" for each basis vector (eigenvector) relates to the probability of measuring the corresponding eigenvalue.$$^*$$ For momentum $$|\psi\rangle=\int|p\rangle\langle p|\psi\rangle\ dp=\int\psi(p)|p\rangle\ dp$$, and we can determine $$\psi(p)$$ if we already know $$\psi(x)=\langle x|\psi\rangle$$: $$\psi(p)=\langle p|\psi\rangle=\int_{-\infty}^{\infty}\langle p|x\rangle\langle x|\psi\rangle\ dx=\frac {1}{\sqrt{2\pi\hbar}}\int_{-\infty}^{\infty}e^{-ipx/\hbar}\psi(x) dx$$

Then $$|\psi(p)|^2$$ can tell us the probability of measuring a momentum value between $$p$$ and $$p+dp$$. For eigenstates of momentum with eigenvalue $$p'$$, $$\psi(p)=\delta(p-p')$$, which means that we can only measure $$p'$$ for momentum.$$^{**}$$

Definite momentum is only achievable with plane waves, which is why my first example worked - but if operators only seem to work on plane waves why do we use them?

We use the operators because the eigenvalues of the operators correspond to the values we can measure of these observables. The operators tell us what we can measure, and they give us nice ways to write our state vectors $$|\psi\rangle$$ so that the probability of measuring these observables are easy to calculate (as shown above).

$$^*$$ If you want something we use to operate to determine the "result" of a measurement, what you want instead is the sum of projection operators. For a general observable with corresponding operator $$A$$, eigenvectors $$|a\rangle$$ and eigenvalues $$a$$, we have $$|\psi\rangle=\sum_a|a\rangle\langle a|\psi\rangle=\sum_a\psi(a)|a\rangle$$ $$|\psi\rangle=\int|a\rangle\langle a|\psi\rangle\ da=\int\psi(a)|a\rangle\ da$$ for discrete and continuous spectra respectively. $$\psi(a)$$ gives you all you need to know about the probability of measuring a value of $$a$$ for our observable.

$$^{**}$$States like these are actually non-physical, but I don't think I need to go into the discussion of this for your question. There are other answers on this site that can go into more detail if you want to learn more.

• I'm afraid I'm having trouble interpreting this due to my lack of experience in this discipline. Firstly, I'm confused with a statement like $\psi(p)=\langle p|\psi\rangle$. To me, the LHS says that $\psi$ is a function of momentum $p$, but equal to the dot/inner product of $p$ with $\psi$. This may well be a misinterpretation, but if this is the case I don't see how this. Why is this the case necessarily? I also don't understand $|\psi\rangle=\int|p\rangle\langle p|\psi\rangle\ dp$, I'm assuming this is an expression of $\psi$ as an integral, which I've never seen before, and (cont) Oct 23, 2018 at 17:57
• @sangstar $|\psi\rangle$ is your state vector. It basically tells you everything about your system in question, kind of like how in classical mechanics everything you need to know for your state is the position and velocity of all of your particles. It can be expressed as a linear combination of any choice of eigenvectors of some Hermitian operator (or you can say $\psi\rangle$ can be expressed in any basis we want). Oct 23, 2018 at 18:02
• (cont) I don't understand conceptually $p$ as a bra vector. Why would an observable be, in itself a vector? Oct 23, 2018 at 18:02
• I understand that it's a state vector, I just don't understand why it's expressed as that integral. I, in my experience so far, usually interpret state vector $\psi$ as a solution to Schrodinger's equation, superposition or eigenstate. Oct 23, 2018 at 18:03
• @sangstar $|p\rangle$ is your momentum eigenvector such that $P|p\rangle=p|p\rangle$. $\langle p|\psi\rangle$ you can think of as a dot product. It tells us "how much of $|p\rangle$ is in $|\psi\rangle$". Therefore, like any vector expressed in some basis: $|\psi\rangle=\sum\langle p|\psi\rangle|p\rangle$. Then we just define $\psi(p)=\langle p|\psi\rangle$, since you can view $\langle p|\psi\rangle$ as a function where if we specify some $p$ we get a number out of it. Oct 23, 2018 at 18:06