1D infinite square well and parity

In summary, the conversation discusses finding solutions to the one-dimensional infinite square well with a potential extending from -a/2 to +a/2 instead of 0 to +a. It also considers the invariance of the potential and wave functions with respect to parity, and discusses the assignment of odd and even parity to the solutions. The individual also shares their approach to solving the Schrodinger equation and their difficulty in finding a solution for the E_1 state. They eventually find a problem with the integration and suggest using a mathematical program for help.
  • #1
siifuthun
24
0
I think I'm on the right track for this problem, but I'm not entirely sure.

Find the solutions to the one-dimensional infinite square well when the potential extends from -a/2 to +a/2 instead of 0 to +a. Is the potential invariant with respect to parity? Are the wave functions? Discuss the assignment off odd and even parity to the solution

So the wave functions should look the same as if it was an infinite square well going from 0 to +a, except it's going from -a/2 to a/2, so the parity should be exactly the same, right? V(x) = infinity at x< -a/2 and when x> a/2.

So I tried solving for the Schrodinger equation using:

http://img145.imageshack.us/img145/4013/03bp1.jpg

And we know that at -a/2 and a/2, the wavefunction must be equal to 0. We also know that for energy level E_1, where n=2, there's a node right at x=0, so the wavefunction must also equal 0 at x = 0 for that case. So:

http://img174.imageshack.us/img174/4408/04ya7.jpg

So we know from above that B=0, plugging that into equation we get:

http://img98.imageshack.us/img98/5520/05ep7.jpg

Now here's my problem, when I try to solve this, I get:

http://img170.imageshack.us/img170/5769/06vs2.jpg

Which is a problem, since it means A=0. Am I wrong in my assumption that the wavefunction for E_1 state is 0 at x=0?
 
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  • #2
There might be a problem with the last statement of E_1 state is 0 at x=0
.
Why do you think that is so?
 
Last edited:
  • #3
Problem with integration

I finally managed to find out what the heck it was. There is a problem with the integration of the function. Try using http://integrals.wolfram.com/index.jsp. Very useful mathematical program.
 

1. What is a 1D infinite square well?

A 1D infinite square well is a theoretical model used in quantum mechanics to describe the behavior of a particle confined to a 1-dimensional region. It consists of an infinitely deep potential well that has a finite width, and the particle is assumed to have no potential energy outside of the well.

2. How does the particle's energy relate to the size of the well in a 1D infinite square well?

The particle's energy levels in a 1D infinite square well are quantized, meaning they can only take on certain discrete values. The size of the well determines the spacing between these energy levels, with larger wells resulting in smaller energy spacings.

3. What is the role of parity in a 1D infinite square well?

Parity is a property that describes the symmetry of a system under spatial inversion, which means switching the coordinates of all particles with their negative values. In a 1D infinite square well, the parity of the wavefunction determines whether it is symmetric or antisymmetric, which in turn affects the allowed energy levels.

4. How does the particle's behavior change with different parity states in a 1D infinite square well?

If the particle's wavefunction has an even (symmetric) parity, it will have an equal probability of being found on either side of the center of the well. On the other hand, if the wavefunction has an odd (antisymmetric) parity, it will have a higher probability of being found closer to the edges of the well.

5. Can a 1D infinite square well have multiple particles in it?

Yes, a 1D infinite square well can accommodate multiple particles, but each particle will have its own set of energy levels and wavefunctions. The Pauli exclusion principle still applies, meaning that no two particles can occupy the same energy level simultaneously.

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