5.1 For a free particle one can set U(x)=0 eV for all x. Consider the state \(e^{ikx}\).
a) Is this a momentum eigenstate? If so, what is its momentum? (its momentum eigenvalue)
b) Is it an energy eigenstate? What is its energy?
c) What is the momentum operator? What is the energy operator in this case? (You can post those here.)
5.2 Consider the wave-function \(\Psi(x,t) = e^{ikx} e^{-i E t/\hbar}\). Suppose k=1nm and E=0.038 eV.
a) Are those consistent with this being the wave-function of an electron? Why or why not? (Please post your thoughts on that here.)
a) Plot this wave-function vs x at t=0. What is its wavelength?
Comment here: In what way is this slightly difficult? How can you resolve that difficulty?
b) Comment on the gif that will appear here pretty soon.
c) Extra credit: Please make a gif that shows plots of this wave function vs x (range about 10 wavelengths). Includes a bunch of frames so that we can see how this wave-function evolves as a function of time. You can post that here or send it to me and I'll post it.
5.3 Consider a wave function which is of the form: \(\psi(x) = A e^{ik_o x} e^{-x^2/2a^2}\) at t=0.
a) I think that the normalization factor is \(A= \frac{1}{\pi^{1/4}a^{1/2}}\). Is that right? Plot \(||psi(x)|^2\).
extra credit: Plot \(\psi(x)\). To do that plot the real part and the imaginary plot as two plots on one graph. Assume k is about 10/a or so. That way there will be lots of wiggles. (added 4-7).
b) Use the KE operator to calculate the expectation value of the kinetic energy at this moment (t=0). (extra credit for posting the operator and the integral here.)
c) Show that you get two terms, and that the cross term is zero. These two non-zero terms should have a fairly simple form. Discuss their interpretation here.
e) extra credit: Assume U(x) =0 and use Fourier analysis to write a time dependent version of this wave-function. (That is, a time dependent version of \(\psi(x) = A e^{ik_o x} e^{-x^2/2a^2}\).)
[Hint: \(\Psi_k (x,t) = e^{ikx} e^{-i E_k t/\hbar}\) is an energy eigenstate for the right value of \(E_k\).] For k= 1 nm and a = 100 nm, calculate \(\Psi(x,t)\) and make a gif showing the evolution of the wave-function as a function of time, and also the evolution of \(|\Psi(x,t)|^2\).
Is this the integral you wanted posted Zack?
ReplyDelete$$
\frac{-A^2 h^2}{2m} \int_{-\infty}^{+\infty} e^{-x^2/a^2} (\frac{x^2}{a^4} - \frac{2ikx}{a^2} - k^2 - \frac{1}{a^2}) dx
$$
I think that you do get the i term when you take a derivative of the e^ikx,,,
DeleteHi Peter and Nicolas. Thanks for those. All your terms look reasonable. They all have the same units. How can we resolve this?
ReplyDeleteGo ahead and post what you get when you integrate as well. It might be valuable to look at that and discuss it.
I agree. The wave-function is complex. Let's do two graphs. One graph for the real part and another for the imaginary part.
ReplyDeleteNICE MAN!!!
ReplyDeleteSo what do those two terms mean? Any ideas on how to interpret and think about them? Maybe we can think about the limits where a is very large, or the opposite limit where k is large . What does each term correspond to?
ReplyDeleteSick!
ReplyDeleteSo I just started the homework because I had other work that I needed to do earlier. Hopefully I'm not too late to the conversation or asking anything that is "dumb."
ReplyDeleteFor 5.1(a), since the potential is 0, it means all that energy would have to be kinetic energy. Since we know that
$$ \hat{K.E} = \frac{\hat{p}^{2}}{2m} $$
The momentum would be Energy over the particle mass.
Is this valid reasoning or is there something I am missing? Also, I am confused what is an energy eigenstate.
Hm, I might be completely wrong about this but in order for a function to be a something-eigenstate, wouldn't we need to satisfy something like:
Delete$$ \psi_{a}(x) = A\psi_a(x) $$
where the right side is the function after placing the operator for something?
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ReplyDeleteFor 5.1(c), the momentum operator would be
ReplyDelete$$ \hat{p} =-i\hbar \frac{\partial}{\partial x} $$
For the kinetic energy operator, it would be
$$ \hat{KE} = \frac{\hbar^{2}}{2m}\frac{\partial^{2}}{\partial x^{2}} $$
sdmsdf,.gmsdfgjoigjaweroigjwalkegtoiejdbl ARGH
Delete$$ \hat{KE} = -\frac{\hbar^{2}}{2m}\frac{\partial^{2}}{\partial x^{2}} $$
Momentum operator:
ReplyDelete\begin{equation}
\hat{p}=-\frac{\hbar}{i}\frac{\partial }{\partial x}
\end{equation}
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ReplyDeleteThe kinetic energy operator: $$
ReplyDelete\hat{T}=\frac{{}\hat{p^{2}}}{2m}=\frac{-\frac{\partial^2 }{\partial x^2}\hbar^{2}}{2m} $$
I think that the integral you're looking for is the following: $$
E(T)=-\frac{A^{2}\hbar^{2}}{2m}\int_{-\infty}^{\infty}e^{-\frac{x^{2}}{a^{2}}}(\frac{x^{2}}{a^{4}}-k^{2}-\frac{2xik}{a^{2}}-\frac{1}{a^{2}})dx
$$
Holy hell, thank you so much. You will save me some trouble. Also, you can use that code and put it into Wolfram|Alpha with a few edits to the text.
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ReplyDeleteCalvin and I agree with everything posted before about 5.3 c) that the two terms are the KE values summed for an HO and unbound particle (the value from 5,1 b). This problem also made me wonder if, in general, multiplying two energy eigenstates and calculating the expectation value will always give the sum of the individual eigenvalues\expectation values.
ReplyDeleteAlso, for the question about considering large and small 'a', we found that in the limit 'a' is small when the particle is very tightly bound the KE is all from the HO contribution and in the limit 'a' is large when the particle is loosely bound the KE is all from the unbound contribution.
We are still confused on the mixed nature of the this answer depending on an expectation value and an eigenvalue if anyone can clarify it would be very appreciated, it seems like an important distinction.
Interesting! I would also very much like to see more discussion and clarification.
DeleteFor 5c there are two non-zero terms in the kinetic energy for this state. To understand their origin it is useful to think about the nature of the wave-function/state. This state is somewhat like a plane wave, due to the \(e^{ikx}\) term, but it is also somewhat localized by the gaussian function. The kinetic energy divides into two terms: one associated with actual motion and one associated with wave-function localization. Does that make sense?
ReplyDelete