Infinite square potential well

  • #1

Homework Statement


I think this is a square well potential problem. The question asks me to sketch the ground-state probability density, for the following situation:

A quasielectron moves in a 'quantum dot' device. The potential V(x) = 0 for 0 ≤ x < L, and is infinite otherwise.

Homework Equations




The Attempt at a Solution


I'm going to need to solve the Schrodinger equation within the device. I have no idea what a quantum dot is, but presumably it doesn't actually matter! I think you can just treat it like an infinite square potential. The Schrodinger equation is

##-\frac{\hbar^2}{2m} \frac{\partial \psi}{\partial x^2} = E \psi##
Within the device. And I also need boundary conditions. One of these boundary conditions is ##\psi(L)=0##, since the potential is infinite at L. Annoyingly, the potential is 0 at x=0, so I can't use the same reasoning there. So my first question is, how do I get my second boundary condition? I don't see how the potential being zero at x=0 in any way restricts what ##\psi## could be...
 

Answers and Replies

  • #2
Vanadium 50
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A. I wouldn't solve anything. It says "sketch".

B. If I were going to solve something, I'd make the problem more symmetric.
 
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  • #3
A. I wouldn't solve anything. It says "sketch".

B. If I were going to solve something, I'd make the problem more symmetric.
Don't I need to calculate the probability density in order to sketch it?
 
  • #4
PeroK
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And I also need boundary conditions. One of these boundary conditions is ##\psi(L)=0##, since the potential is infinite at L. Annoyingly, the potential is 0 at x=0, so I can't use the same reasoning there. So my first question is, how do I get my second boundary condition? I don't see how the potential being zero at x=0 in any way restricts what ##\psi## could be...
Specifying ##V(x) = 0 \ (0 \le x < 1)## is equivalent to ##V(x) = 0 \ (0 < x < 1)##.
 
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  • #5
Specifying ##V(x) = 0 \ (0 \le x < 1)## is equivalent to ##V(x) = 0 \ (0 < x < 1)##.
Why? In the first inequality V(x) is zero at x = 0, but in the second inequality the potential is infinite at x = 0. So I can't say that at x = 0, the wavefunction is zero, and I don't know where else to get a second boundary condition from.
 
  • #6
PeroK
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Why? In the first inequality V(x) is zero at x = 0, but in the second inequality the potential is infinite at x = 0. So I can't say that at x = 0, the wavefunction is zero, and I don't know where else to get a second boundary condition from.
Whether the boundary points are in or out doesn't affect the functions or the integrals. If the wavefunction is not ##0## at ##x=0## then it is discontinuous when viewed as a function for all ##x##.
 
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  • #7
Whether the boundary points are in or out doesn't affect the functions or the integrals. If the wavefunction is not ##0## at ##x=0## then it is discontinuous when viewed as a function for all ##x##.
So if I said the wavefunction is zero at x = 0, then I'm excluding the electron from being at x = 0, even though at x = 0 the potential is zero... that's ok? I thought it would have something to do with continuity! I was trying to find a way to use the derivative of ##\psi##.

Vanadium's post makes it sound like I don't actually need to solve for ##\psi## though to sketch the probability distribution. I know it's zero at the edges, but don't I need to integrate the magnitude between infinity and negative infinity to work out the probability density? How could I sketch it without an equation?
 
  • #8
PeroK
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So if I said the wavefunction is zero at x = 0, then I'm excluding the electron from being at x = 0, even though at x = 0 the potential is zero... that's ok? I thought it would have something to do with continuity! I was trying to find a way to use the derivative of ##\psi##.

Vanadium's post makes it sound like I don't actually need to solve for ##\psi## though to sketch the probability distribution. I know it's zero at the edges, but don't I need to integrate the magnitude between infinity and negative infinity to work out the probability density? How could I sketch it without an equation?
Yes. I suspect whover set the question didn't give it a second thought!
 
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  • #9
Yes. I suspect whover set the question didn't give it a second thought!
I know I'd only actually have to integrate between L and 0, and it should equal 1. And it's zero at both edges. Where does the whole 'ground state' thing come into it, anyway?

If this is actually a really easy question, I must be missing some fairly basic information.
 
  • #10
PeroK
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I know I'd only actually have to integrate between L and 0, and it should equal 1. And it's zero at both edges. Where does the whole 'ground state' thing come into it, anyway?
Are you supposed to know the ground state of the infinite square well? If not, then to sketch it, you'd have to solve the Schrodinger equation, I imagine.
 
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  • #11
Are you supposed to know the ground state of the infinite square well? If not, then to sketch it, you'd have to solve the Schrodinger equation, I imagine.
OK. We haven't done anything specifically about ground states of infinite square wells. I'd better solve the Schrodinger equation then!
 
  • #12
PeroK
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OK. We haven't done anything specifically about ground states of infinite square wells. I'd better solve the Schrodinger equation then!
It can't do any harm!
 
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