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I have been recently studying QM and I have encountered the case of a free particle. I understood that a free particle travels in the form of a wave packet where we get $$\psi (x) = \frac{\int_{-\infty}^{\infty} \phi (k) e^{ikx} dk}{\sqrt {2 \pi}}$$ Now given $\psi (x,0)$ I can find $\psi (x,t)$ now we are required to find the probability density of the wave function. I think it should be $\psi ^*(x,t)\psi (x,t)$ but the solution given does $\phi^*(k,t)\phi(k,t)$ and I am confused. Added to this, we are also given momentum of the particle is $2\hbar k$ but I have no idea where we need to use this value or if it has any significance or not. Also we are required to find the mean energy and I did the normal way: $\int_{-\infty}^{\infty}\psi^* (x) H \psi (x)$ but the answer does not match. Am I interpreting something wrong about the free particle? Please help, I am in real confusion would be glad for some hint.

I found this link where it says $\phi(k)$ is the probability amplitude of momentum of the free particle, but wont we find the expectation value of momentum of the particle by this formula $$\int_{-\infty}^{\infty}\psi^* (x) \frac{-i\hbar d}{dx} \psi (x)~?$$ Probability density for momentum in Quantum Mechanics

To be precise: I am posting the question statement here:

At time =0, a free particle in quantum mechanical state is described by the wave function ()=$\sqrt{\frac{\alpha}{\pi}}e^{\frac{-\alpha x^2}{2}}$.

(a) Find the probability density of the particle with momentum 2ℏ at any time t. Here, k is the wave vector.

(b) Find the mean energy of the particle at any time t.

Note: This is not a HW question. Rather a question that came in our college exam.

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    This is not a HW question. That’s not relevant. Whether or not a question is considered homework-like on this site has nothing to do with whether it is or was actual homework for you or anyone else, or whether it is or was on some exam for you or anyone else. The decision is supposed to be based only on what kind of question you are asking. – G. Smith May 20 '21 at 19:47
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    Have you translated the question statement into English? It seems like it should be asking “Find the probability density for the particle to have momentum 2ℏ at any time t.” This would be a probability density in momentum space. – G. Smith May 20 '21 at 19:52
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    You can write $\hbar$ as \hbar. – G. Smith May 20 '21 at 19:55
  • Your expectation value of momentum appears to lack a differential. – Gert May 20 '21 at 19:59
  • gaussian wave packet There is nothing Gaussian about what you have written. – G. Smith May 20 '21 at 19:59
  • @G.Smith I have not translated the statement. It was originally in English only. And Gaussian wave packet, I used this word since I had read the wave packets are mostly Gaussian in nature. I may be wrong. Should I edit it? –  May 20 '21 at 20:20
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    What you wrote is a Fourier transform. It can represent a wavepacket with any shape, not just a Gaussian shape, so I recommend removing the word “gaussian”. – G. Smith May 20 '21 at 20:34
  • You seem to have some confusion between probability and expectation. I also fear that you're trying to memorize formulae ("I did the normal way..."). There are several good books on quantum theory that really help you understand and that explain the concepts step by step. Take a look for example at de Muynck's Foundations of Quantum Mechanics and Peres's Quantum Theory: Concepts and Methods. Check the formalism of POVMs, it makes measurement & probability much clearer. – pglpm May 22 '21 at 20:25
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    the definition of $\psi(x)$ as an integral over $dx$ is wrong. You are supposed to be coherently adding momentum eigenstates $\exp{ikx}$ with momentum on $(k,k+dx)$ with weight $\phi(k)$. – JEB May 22 '21 at 20:28
  • @JEB, yep, you are right, that was a typo. fixed it. But how did you know momentum eigenstates are exp ikx and momentum eigen values are $\phi(k)$ ? like I have read Griffith cover to cover but the book didn't mention what is $\phi(k)$ , so what I interpreted was that those are the coefficients we generally have in case of Asinkx+Bcoskx , A and B kinds, and I cant link this with momentum at all. –  May 22 '21 at 22:04
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    In order to address all the confusions you have, it would seem necessary to write the introductory chapters of a QM text. Even Griffiths contains the answers to these questions, which you say you have read cover to cover. There are many other QM texts (and online notes) if you found Griffith's presentation unclear, though few would be as mathematically gentle. – Richard Myers May 22 '21 at 22:33
  • Yes, I was a bit confused about finite square well potential too and I found MIT courseware useful but regarding free particle, I couldn't find any good source (open-source), could you please share some open source material that discusses my doubts preferably something that has a thorough discussion on free particle. –  May 22 '21 at 22:42
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    @Alex: $-i\partial_xe^{ikx}=ke^{ikx}$ so $e^{ikx}$ is an eigenstate with momentum $k$. – JEB May 22 '21 at 23:22
  • This is a Gaussian wave packet. It should be featured in your text, and most decent QM texts. Read up. – Cosmas Zachos May 23 '21 at 00:46

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Let me provide you some insights into your problem.

  1. Why are we doing $\phi(k, t)^*\phi(k, t)$ instead of doing the normal $\psi(x, t)^*\psi(x, t)$? This is simply because $\psi(x, t)^*\psi(x, t)$ represents the probability density of particle being found in a given position eigenstate. In simpler terms, being found at a particular position in the one- dimensional space. But, if you read your question carefully, it is asking you to find the probability of being found in a momentum eigenstate. So, we jump ships to the momentum space. Here, is where the Fourier transform and Inverse Fourier tricks jump in. Recall the postulates of Quantum mechanics and you will find that the square of the coefficients in the expansion of an eigenfunction in terms of the eigenvectors of an observable represent the probability of being found in that eigenstate of the observable. So, doing $\phi(k, t)^*\phi(k, t)$ is correct apart from some inherent constants which you can find out easily if you look up Fourier transform in any standard text.

  2. This is much more trivial and easy to find out if you look carefully. Your wavefunction is not normalized so your formula will not work. Normalize first and then try using that formula or simply do $$\langle E\rangle=\frac{\int_{-\infty}^{\infty}\psi^*\hat H\psi dx}{\int_{-\infty}^{\infty}\psi^*\psi dx}$$

Manas Choudhary
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