Calculating the expectation value for a particles energy in a 1_D well

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expectation value for a particle in a 1-D well

how do i calculate the expectation value for the particles energy in a 1-D well.

i have attached a word file, with my working out, just not quite sure if I am on the right track...

i appreciate any help...thanks a mill
 

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Erm, unless it means something else, L_z is an operator, not something you'd find in a wave function. So that wave function doesn't really make sense.
 
I can't really read that document at all, the equations are extremely blurry and I can't make out the symbols. However, recall that the expectation value for a given operator is the following:

<br /> \langle Q_{op}\rangle = \int_{-\infty}^{\infty} \Psi^{*} Q_{op} \Psi dx<br />
 


can you explain to me what you did not understand from the question or my working out...so i can clear it up..

Lz is not an operator, its just part of the function to describe the the particle in a 1D well.
 


have a hand written version attached..
 

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Your calculation of ##\hat{H}\psi## isn't correct in both of your attempts. I'm not sure how you're getting your result. Do you know how to calculate a derivative?
 
i know how to calculate the derivative, can you please guide me.

thanks a mill
 
If you know how to differentiate, can you explain how you got your results?
 
Im not sure if I am doing it right,
what are the steps of solving this problem,

1st. find a eigen value of hamiltonian
2. find the time dependence wave function for stationary states
am i right..

can you please guide me through it
 

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  • #10
As Clever-Name implied above, you want to calculate ##\langle \psi \lvert \hat{H} \rvert \psi \rangle##, so calculating ##\hat{H}\psi(z)## is a good first step. You have
$$\hat{H}\psi(z) = -\frac{\hbar^2}{2m}\frac{d^2}{dz^2} \sqrt{\frac{10}{L_z}}\left(1-\frac{z^2}{L_z^2}\right).$$ So far so good. The problem is, the rest of what you wrote doesn't make sense. First, you claimed
$$-\frac{\hbar^2}{2m}\frac{d^2}{dz^2} \sqrt{\frac{10}{L_z}}\left(1-\frac{z^2}{L_z^2}\right) \rightarrow -\frac{\hbar^2}{2m}\frac{n^2\pi^2}{L_z}\left(1-\frac{z^2}{L_z^2}\right)$$ which is simply wrong. This is basic differentiation. I don't understand how you got your result if you know how to calculate a derivative. Next, you then somehow claim that your answer is of the form
$$E\sqrt{\frac{10}{L_z}}\left(-\frac{z^2}{L_z^2}\right)$$ which is obviously inconsistent with what you wrote earlier (what happened to the 1 inside the parentheses?). In fact, the relationship ##\hat{H}\psi = E\psi## only holds if ##\psi## is an eigenstate of the Hamiltonian, and the given wave function isn't an eigenstate.

You never said what the actual potential here is. I'm assuming you have an infinite square well as opposed to a finite square well.
 
  • #11
i have attached a file with integration of wavefunction.

so i integrated the wavefunction, I am just not sure of the steps.

for example> if i want to work out the expectation value of energy. what exact steps do i follow..

im just confused with the terminology of what I am asked.
 

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