Potential well, limits changed, quantum

[Chronos000]

Homework Statement

If you have a 1d infinite potential well between 0 and a which is then expanded to become 0 -2a with the wavefunction in that instant undisturbed, what is the expectation value of energy at that moment.

my thoughts are that as the eigenfunction will be the same then so will the eigenvalues. so therefore the energy will just be the same as what it was before. I just need this confirmed.

 

[ideasrule]

You have the right answer, but for the wrong reason. The eigenfunctions won't remain the same--one additional stationary state, namely the ground state, is introduced when the well expands. Also, the coefficients of the stationary states in the wavefunction will change drastically.

The expectation value of energy stays the same simply because energy is conserved.

 

[Chronos000]

Thanks for your reply, can you explain further how an additional ground state can be introduced? I thought for the ground state you have n=1 and that was that.

 

[ideasrule]

I meant that the new ground state did not previously exist. The n=1 state is always Asin(n*pi*x/L), and your L is now twice as large. Not only that, none of the new n=3,5,7,9... states previously existed.

 

[L-x]

It may help you to think about what's happening if you imagine the particle is in the ground state to start with, and to draw it's wavefunction.

[PLAIN]http://ursula.chem.yale.edu/~batista/vvv/img230.png

now if you move the right hand vertical line to the point x=2a the wavefunction will still have the same shape, but will now be made up of (in this case) an infinite sum of different amplitudes of the eigenfunctions of the new box. If you have trouble with this idea it might be helpful to revise Fourier series.

edit: It should be obvious that the above reasoning holds from some general wavefunction, it would just be slightly less easy to draw.

 

[Chronos000]

I think I understand this now thanks

 

[Chronos000]

while we are on the subject actually, if you are asked to solve a 1d infinite pot well between -a and a. are you allowed to shift it to between 0 and 2a. from what you have said above it would seem like the wavefunction would still be the same

 

[L-x]

the shape of the wavefunction is independent of what you call your co-ordinates, so yes it would be the same shape but it would be in a different place and therefore it would not be the same function (sin x is just cos x in a different place). you would effectively just be redefining x->x+a, which, of course, you are allowed to do (you could make the potential well between 100 and 100+a if it pleased you) be aware however that as you have redefined x you must (re)redefine it back again, and adjust your function accordingly, if you wish to present your answer in the original co-ordinate system. It seems like a lot of bother for no real gain to me.