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  The evolution of a one-dimensional wave packet from exercise 3.3 of Cohen's Quantum Mechanic.

+ 1 like - 0 dislike
105 views

Hi everyone,

I have some doubts in a exercise from chapter 3 of Quantum Mechanics (Cohen-Tannoudji). It is number 3 from that chapter. It defines a wave function of a free particle at time t = 0 as:

\[\phi (x,0)\ =\ N\int_{-\infty}^{\infty}e^{{-\mid k \mid}/k_o} e^{i k x} dk\]

where \(k_o\)and N are constants.

It first asks for the probability of a measurement of the momentum, performed at time t = 0, which yields a result included between \(-p_1\ \)and \(p_1\). That is easily done.

Then, it asks the same for a measurement performed at time t. Now, I arrived at the same probability, which makes sense because it is a free particle.

After that, it asks for the form of the wave packet at time t=0. And that is the following:

\[\phi (x,0)\ =\ \frac{2N/k_o}{{1/k_o}^2+x^2}\]

I wanted to show the plot but for some reason I was not able to upload it. Its shape as similar to a gaussian wave packet. It then asks to calculate \(\Delta X \Delta P\ \)for that time t which yields \(\hbar/2\). Finally it asks to describe qualitatively the wave packet evolution in time. My reasoning is the following:

As it looks similar to a gaussian wave packet, a first guess would be that it behaves similarly as well. Therefore, the wave packet should spread as it evolves through time. That means \(\Delta X\ increases\ and\ \Delta P\ decreases.\) But if that is true, then there probability  of our measurement should increase also and not stay constant.

I would be really grateful if anyone could elucidate this doubt.

Best Regards,

Frantic Undergrad

asked Jun 5, 2018 in Theoretical Physics by anonymous [ no revision ]

1 Answer

+ 1 like - 0 dislike

Interisting question.

Your doubt arives from the fact that you think that in the Guassian wave packet uncertanty in momentum ΔP decreases over time. That's not true.

As the Guassian wave packet propagate the position wave function does spread and therefore uncertanty in position ΔX does increases. But ΔP does not, it simply don't, just look at the equation for the momentum wave function and you will notice that it is a guassian function with fixed mean square deviation for all times. Meaning ΔP does not change at all over time. It can be shown (as Cohen's next exercise almost asks you) that ΔP is constant for all times for a free particle, only ΔX changes. This already constitues a rigorous demonstration for you to say that the probabilty that you calculated before won't change. Try doing it, it's not difficult.

You might be thinking, what about Heisenberg uncertanty principle? Shouldn't "ΔPΔX>=constant"?????

Well Heisenberg uncertanty principle never said the product ΔPΔX should be constant over time, it only says it should be equal OR greater the constant for all times. What happens to the guassian wave packet is that the product ΔPΔX increases over time, with ΔX increasing over time and ΔP remaining constant.

These links will provide you all the visualization you need for that. (please visit them, they were life-changing experiences in my physics learning experience):


https://www.cond-mat.de/teaching/QM/JSim/wpack.html

So in Cohen's question it describes a very similar situation to a Gaussian wave function that does not propagate. Its position wave function will spread but it will always be centered around the origin. That happens cause the average momenta is zero (as you calculated). The momentum wave function will not spread, not changing ΔP and therefore not changing the probability that you calculated before, as you should expect for any free particle. 

There is of course a more difficult way of showing it, but it requires you to calculate the position (or momentum) wave function for all times. That can be more easily done through use of the Feynman path integral formalism as described here (just change the guassian packet for Cohen's function):
https://physics.stackexchange.com/questions/386332/analytical-solution-for-a-gaussian-wave-packet-free-particle
With that in hands you could calculate ΔP for all times and realize it is constant over time. The integrals are not easily done.

Best Regards 

Pedro Dardengo.

P.S.: I would like a more detailed discussion of question 11 of chapter 3, if you happen to know some material about it, please tell me.

answered Feb 7 by pedrodardengo (10 points) [ no revision ]

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