**Required math: calculus **

**Required physics: **Schrödinger equation

Reference: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 2.5.

As an example of an explicit case of a particle in the infinite square well, suppose we have a particle that starts off in a combination of the two lowest states:

To normalize, we find by using the orthonormal property of the stationary states, so:

So .

Using , we have for the full wave function:

and

The stationary states are real functions, so we can drop the * notation on the complex conjugates. Note that for all times, since the cosine term integrates to zero due to orthogonality.

The average position is:

The particle’s mean position oscillates about the midpoint of the well with an amplitude of .

The mean momentum can be found the quick way by taking the derivative of .

where we have used the definition of . Doing it the long way using integration does give the same answer, as can be checked using Maple (or by hand).

The two possible energies are and and since the wave function consists of equal contributions from the corresponding stationary states, they occur with equal probability. Thus

Again, this can be obtained the long way through integration.

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Rocky WongThis line was wrong, there is an extra factor of 2 in front which should not be there.

displaystyle 2|Psi(x,t)|^{2} displaystyle = displaystyle (psi_{1}^{*}(x)e^{iomega t}+psi_{2}^{*}(x)e^{4iomega t})(psi_{1}(x)e^{-iomega t}+psi_{2}(x)e^{-4iomega t}).

Reference:

http://www.lns.cornell.edu/~dlr/teaching/p443/ps1_solutions.pdf

growescienceIf you mean equation 7, I believe it’s correct. It leads to equation 9 which agrees with the solution in your reference (the equation below your equation 5).

AnonymousOh yes I got it! It was a silly algebra mistake by me. Sorry for that!

Cindyyou describe equation 9 by this

” since the cosine term integrates to zero due to orthogonality”

but i don’t get it @@ cosine function has this property!!??

i thought that it’s equal 1 just because the normalization of wavefunction

gwrowePost authorThe functions and in the cosine term are orthogonal, not the cosine.

cindyWhy should we find Cn using eq.37 in example2 but we can just put on the time dependent term in (6) in this problem??

And I tried to find Cn using eq37 and found it to be zero@@???

gwrowePost authorDon’t understand your question – there is no Cn or equation 37 in this post.

cindyWell the equation 37 I mean is that in this book noted as 37

Using “fourier’s trick” to get the coefficient at arbitrary given time at first

And the example I mean is also the book’s example2 in chapter2

DavidI am confused on where the sqrt(2)/2 comes from in eqn 6. Is that Cn? If so how is it calculated. Thanks for your help and your solutions are a great companion to the book.

gwrowePost authorIt’s just as calculated in equations 2 through 5.

Kaywhere did the integration come from for equation 10 to 11? was an example used from anywhere in the book to help with this?

gwrowePost authorI’ve plugged in the stationary state wave function

I did the integrals using Maple, but if you want to do them by hand they are fairly standard integrals – you’d need to use a trig identity or two and integration by parts (or you could just look them up in tables).

Sun Bowhat means “equal contributions from the corresponding stationary states”

from which term can we judge that

gwrowePost authorBoth stationary states and have the same coefficient in equation 1, so they have equal contributions to the overall wave function.