These questions are not designed to be clear, so it is natural to want to ask questions about them and discuss them. Please read and engage with them as best you can and ask questions and join in the discussion here. Part of your learning is to figure out what is being ask here, what these things mean, and what is expected. I view posting here as a sign of engagement and I appreciate that.
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Added Note:
How functions work: Consider the function f(x) = x^2. What is the nature of the related function f(x-9a)?
At x= 9a, what is f(x-9a)? zero right? And at x=10a one would get a^2, I think. True?
It is no different with our function, v(x) above. Whenever the argument is zero, the rules of the function yield v=-10 eV;
whenever the argument is greater than L/2 then the function is zero. For the function v(x-2a) the argument is zero when x=2a. For the function v(x+2a) the argument is zero when x=-2a. Although it could be something in the way I worded the problem or defined the function, I think in this case the issue is that you have not been taught clearly what a function is. If in resolving this you acquire a clearer concept of that, that would be the best outcome. The argument of the function is the key thing.
well, you have to square it first. It is a wave function so the normalization condition is on it after squaring it.
ReplyDeleteThen it is probably easiest to integrate from zero to infinity (to get rid of the abs value) and then double that. does that make sense?
That is a key question!
ReplyDeleteWhat is b?
What is a?
In the physically relevant region for solids and molecules, in general, and semiconductors, in particular, how are b and a typically related?
how do we graph V(x) = sum U(x-ja) when U(x) is a constant and not a function of x?
ReplyDeletewill it simply be a few sq. wells that are in series?
DeleteThis sounds correct. Another way to think of this is that you are just shifting the same well by a factor with each iteration. This will give you multiple wells in a particular arrangement.
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ReplyDeleteFor problem 3, is x such -L/2< x <L/2?
ReplyDeleteAs I understand, problem 3 is a bit different... try sketching a single iteration of the function i.e for n=0. What does it look like?
DeleteFor problem number 3, inside the psi function, is the phi a function of (x-na) or multiplying by (x-na)?
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DeleteTo me what your comment suggests is that perhaps we need to be more clear on what a function is and how functions are defined and implemented. For example, given the definition of a function v(x), as above, how is v(x-10a) different from v(x) and how is it similar?
Deletephi is a function of (x-na). It is definitely NOT multiplying by (x-na).
Deletewell, a >> L so since v is dependent on U and U is 0 other than between -L/2 and L/2. Thus v(x-10a) is 0 everywhere. Is that correct?
DeleteNot correct. v(argument) is a function. When its argument is 0, then the function is -10 eV. When its argument is greater than L/2, then the function is zero. That is the rules of the function. That is how functions work.
DeleteWould you understand this better if the function were x^2 ? I will edit a comment on functions into the post above.
It's not multiplied. Shifted. Argument change.
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ReplyDeleteHey John, you might remember me from every class and conversation we've shared this year. Do you think it's significant that it's doubly infinite, as opposed to infinite in one direction? How do you think it would change the behavior of the electron, if at all? (Note: My deleted comment is the same as this one, I just wanted to see if I could figure out how to add a profile picture).
ReplyDeleteSo for 1c, are we graphing how much each well differs from the original well where j=0?
ReplyDeleteNo. It means sum on j from -2 to +2, but skip j=0.
Deleteso then you have 4 terms: j=-2,-1, 1, 2
(instead of the 5 terms you have in b))
Isn't the sum equal to zero for part 1c?
DeleteOnly the j=0 yields a quantity to be added for part 1b.
No. v(argument) is a function. when its argument is 0, then the function is -10 eV. When its argument is greater than L/2, then the function is zero. That is the rules of the function. That is how functions work.
DeleteI am just using my mathematical intuition. I hope it is correct. :)
ReplyDeletePlease don't post as anonymous. You can use a made-up name if you like.
DeleteI made an email account.
DeleteLol at your name
DeleteFor the extra credit part of question 3 would it be reasonable to create two different graphs, one in the real, and one in the complex plane for our answer?
ReplyDeleteyes! excellent point. It would be the only way that I can see.
DeleteU(x) = -10eV $\frac{L}{2} < x < \frac{L}{2}$
ReplyDeleteNotice that this is not the same x as the x in V(x). V(x) is a superposition of U with different parameters. Since it's a trivially small summation, you can write it out like:
V(x) = U(x+2a) + U(x+a) + U(x) + U(x-a) + U(x-2a).
Now you can clearly see that V(x) is the superposition of 5 U functions with different shifts along the x axis. The arguments for x in U(x) are the shifted values, which tells you that the shift corresponds to the center of the well since that is what the $\frac{L}{2} < x < \frac{L}{2}$ corresponds to.