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Is $\langle\psi_1|p\psi_1\rangle$ necessarily 0 for harmonic oscillator eigenstates?

If $\Psi(x,t)= c_0\psi_0(x)e^{-iE_0t/\hbar}+c_1\psi_1(x)e^{-iE_1t/\hbar}$,

is the following true? Where $p$ is the momentum operator,

$$\langle\psi_1|p\psi_1\rangle = \langle\psi_0|p\psi_0\rangle=0.$$

Any hint would be appreciated.

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closed as off-topic by Abhimanyu Pallavi Sudhir, John Rennie, Emilio Pisanty, Waffle's Crazy Peanut, WetSavannaAnimal Nov 11 '13 at 4:58

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  • $\begingroup$ hint: the r in \ranlge stands for right. Also, that bracket will generally not be zero. Otherwise the function must be constant w.r.t. $x$ which, for normalizations, implies your volume must be finite. $\endgroup$ – Nikolaj-K Nov 6 '13 at 23:24
  • $\begingroup$ Do you mean eigen states of the Hamiltonian operator? Besides I think it certainly depends in the representation you're using, remember that there is Schrödinger picture, in which the kets evolve with time, the Heisenberg picture in which kets remain stationary while the operators evolve with time, and lastly there is an Interaction picture. $\endgroup$ – Oscar David Arbeláez Nov 6 '13 at 23:26
  • $\begingroup$ @OscarDavidArbeláez: The wave function $\Psi(x,t)$ he wrote down has an explicit time-dependence, so I think we can assume he's using the Schroedinger picture. Also, since $\Psi(x,t)$ is a superposition of terms on the form $\psi_i(x) \exp(-iE_i/\hbar)$, the $\psi_i$ should be the Hamiltonian eigenstates. $\endgroup$ – jabirali Nov 7 '13 at 1:48
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Which system are you talking about? I believe this is not true in general...

However, eigenstates of all the simplest quantum mechanical systems (free particle, harmonic oscillator, hydrogen atom etc) have definite parity. Suppose the $\lvert\psi_1\rangle$ you are talking about has $$P\lvert\psi_1\rangle=+1\lvert\psi_1\rangle$$ Also, in the coordinate representation, the momentum operator is $$\hat{p}_x=-i\partial_x$$ which being a derivative changes the parity of the state to which it is applied. $$P\lvert p\psi_1\rangle=-1\lvert p\psi_1\rangle$$

Therefore you are projecting a state of definite parity onto a state of opposite definite parity.

That is why you have a $0$, maybe...

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  • $\begingroup$ You should mention that eigenvectors of Hermitian matrices are orthogonal, and that the parity operator is Hermitian. $\endgroup$ – lionelbrits Nov 6 '13 at 23:48
  • $\begingroup$ You are right. I gave that as a known fact, but yes, it should be said in order to make better sense of the answer. $\endgroup$ – Federico Carta Nov 7 '13 at 0:47
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Recall that $p = i\sqrt{\frac{m \omega\hbar}{2}}(a^{\dagger}-a)$. The result follows immediately. Although @FredericoCarta 's method is perhaps more insightful.

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As @FredericoCarta mentionned, you need to be more precise with your system, especially with the momentum operator $P$ which is very tricky as it is hermitian, but not necessarily self adjoint depending on its domain of definition $\mathcal D(P)$.

Hence, you may find a state $|\Psi\rangle$, that is indeed eingenstate of the momentum operator, (for example $e^{i p x}$) but which doesn't belong to your initial Hilbert space, hence you have $p=0$.

Note that this state then usually belongs to $\mathcal D(P^\dagger)$), as $\mathcal D(P)\subset \mathcal D(P^\dagger)$)

Hope this helps.

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For the harmonic oscillator, $\langle p\rangle$ is always going to be $0$ for a single energy eigenstate. An energy eigenstate is a "stationary state", which means that only the phase of the wavefunction changes with time. Consider what would happen if $\langle p\rangle$ were positive or negative: the wavefunction would be more likely to be moving to the right (positive) or left (negative). This would result in movement of the average value of $x$, but $\langle x \rangle$ can't change because you're in a stationary state.

Another way to look at it is that the hamiltonian is symmetric in both $x$ and $p$. If $p$ were nonzero and pointed in one direction or another, what broke the symmetry?

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In general, the expectation value of momentum for any non-relativistic, eigen-, bound-state is zero, which includes the harmonic oscillator eigenstates. Since at one hand $$ \langle n | [x, H] | n \rangle = \langle n | [x, \frac{p^2}{2m}] | n \rangle = \frac{1}{m} \langle n | [x, p] p | n \rangle = \frac{i}{m} \langle n| p |n \rangle \tag{1} $$

on the other hand, $$ \langle n | [x, H] | n \rangle = \langle n | x H - Hx| n \rangle = \langle n | x H |n \rangle- \langle n| Hx| n \rangle = E_n \langle n|x| n \rangle - E_n \langle n|x| n \rangle \tag{2} $$

Since for bound state, $\langle n | x |n \rangle$ is a well-defined number, the right-hand-side of Eq (2) is zero.

The requirement of bound state is necessary. Since (i) the expectation value of momentum for a momentum eigenstate is in-general, non-zero (we are answering what is the momentum for a momentum eigenstate); (ii) for momentum eigenstate, $\langle x \rangle $ is not a well-defined number, which corresponds to the average position for a plane wave...

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