0
$\begingroup$

The question is as follows,

Consider a simple harmonic oscillator in its ground state. An instantaneous force imparts momentum $p_0$ to the system. What is the probability that the system will stay in its ground state.

My approach to the problem is as follows,

Since the Hamiltonian and the momentum operators don't commute (due to the presence of a non zero $V(x)$ term in the Hamiltonian), the energy and the momentum eigenstates are different. And now we know that the system has a definite momentum $p_0$, hence the system is now in the state $\vert p_0 \rangle$. The position representation of this state is given by $\vert p_0 \rangle \, \dot{=} \, \frac{1}{\sqrt{2\pi\hbar}} e^{\frac{i}{\hbar}p_0x}$, and we are interested in the probability, $\vert \langle \phi_0 \vert p_0 \rangle \vert^2$, where $\phi_0$ is the ground state of the quantum harmonic oscillator. This gives us the integral, $$\mathcal{P}_0 = \left\vert \int_{-\infty}^{\infty} dx \left(\frac{m\omega}{\pi \hbar}\right)^{\frac{1}{4}} e^{-\frac{m\omega}{2\hbar}x^2} \cdot \frac{1}{\sqrt{2\pi\hbar}} e^{\frac{i}{\hbar}p_0x} \right\vert^2 = \frac{1}{\sqrt{m\omega\pi\hbar}}e^{-\frac{p_0^2}{m\omega\hbar}}$$

Well this was all good until I checked the solution, where it has been claimed that the state of the quantum system right after the momentum is imparted is given by $$\vert p_0 \rangle = e^{-\frac{i}{\hbar}p_0x}\cdot \vert0\rangle$$ and they proceed with the probability calculations. I am not very sure how this result is true. Any help is appreciated.

$\endgroup$
5
  • $\begingroup$ It's not clear what your difficulty is. Is it the minus sign? Or is it the ket? I don't know if this is relevant to your question, but you do have an error in your question, Your expression for the position representation of the momentum isn't. What you have on the RHS is really $<x|p_0> = \exp (i/\hbar p_0 x)$ $\endgroup$
    – garyp
    Sep 7 '20 at 14:02
  • $\begingroup$ @garyp My difficulty is in the ket. If you notice, I have directly used $\langle x \vert p_0 \rangle$ in my probability calculations. But in the actual solution they use $\langle x \vert p_0 \rangle = e^{-ip_0x/\hbar} \langle x \vert 0 \rangle$. Is there anything wrong with my understanding? $\endgroup$ Sep 7 '20 at 14:12
  • $\begingroup$ You are saying "we know the system has definite momentum". This is different from the question which says "an instantaneous force imports momentum". That means, that we just add that momentum to our particle and not that we are in a momentum eigenstate. $\endgroup$ Sep 7 '20 at 15:28
  • $\begingroup$ @MartinPeschel Thank you for your reply. In such cases, we just multiply the momentum eigenstate with the initial state? Can you suggest any material that I could refer to, for this? Also, I assumed it's in the state $\vert p_0 \rangle$ because, like I have mentioned the momentum operator doesn't commute with the Hamiltonian, and hence if we know that $p_0$ is imparted for sure, is it not equilvalent to saying the system is in the state $\vert p_0 \rangle$? I mean, if we try measuring the momentum (hypothetically), would we not find it to be $p_0$ with absolute certainty? $\endgroup$ Sep 7 '20 at 15:32
  • $\begingroup$ I think Kaper answered this $\endgroup$ Sep 7 '20 at 16:10
2
$\begingroup$

The phrase "An instantaneous force imparts momentum $p_0$ to the system." is a bit ambiguous, the entire concept of a force is not well-defined in quantum mechanics in general. So what the phrase exactly means is somewhat open to interpretation, and your interpretation differs from the one by the author of the solutions.

Before the impulse happened, the system is in a certain state in momentum space, $\phi_0(p) = \langle p | 0 \rangle$. While not an eigenstate, this wavefunction defines a probability distribution on what momentum you could measure the system to have. You assume that after the impulse, all that is forgotten and the system is now in an eigenstate with momentum $p_0$. That's a bit odd. Suppose in classical mechanics you were told a force was exerted on a moving particle, wouldn't you add the given momentum to the momentum the particle already had? That's what the author assumes, that the momentum is added to the system. They then take that to mean the wavefunction is translated in momentum space, which in coordinate space corresponds to a multiplication by $e^{\frac{i}{\hbar} p_0 x}$. Notice that $$\hat{p} \left(e^{\frac{i}{\hbar} p_0 x}|0\rangle \right) = \frac{\hbar}{i} \frac{d}{dx}\left(e^{\frac{i}{\hbar} p_0 x}|0\rangle \right) = p_0 e^{\frac{i}{\hbar} p_0 x}|0\rangle + e^{\frac{i}{\hbar} p_0 x}\frac{\hbar}{i}\frac{d}{dx}|0\rangle = e^{\frac{i}{\hbar} p_0 x} (\hat{p} + p_0) |0\rangle $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.