What Are the Solutions to the 1D Infinite Square Well from -a/2 to +a/2?

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SUMMARY

The discussion focuses on solving the one-dimensional infinite square well potential extending from -a/2 to +a/2. The wave functions exhibit parity symmetry similar to the well defined from 0 to +a, with the potential being infinite outside the boundaries. The Schrödinger equation is employed to derive the wave functions, revealing that for the energy level E_1 (n=2), the wave function must equal zero at x=0, indicating a node. A critical issue arises from integration errors, which can be resolved using the Wolfram Integrals tool for accurate calculations.

PREREQUISITES
  • Understanding of quantum mechanics principles, specifically the Schrödinger equation.
  • Familiarity with the concept of infinite square wells in quantum physics.
  • Knowledge of parity and its implications on wave functions.
  • Proficiency in integration techniques relevant to quantum mechanics problems.
NEXT STEPS
  • Explore the implications of parity in quantum mechanics and its effect on wave functions.
  • Learn about solving the Schrödinger equation for various potential shapes, including finite wells.
  • Investigate the use of computational tools like Wolfram Integrals for solving complex integrals in physics.
  • Study the properties of wave functions in quantum systems, focusing on nodes and energy levels.
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Students and professionals in physics, particularly those specializing in quantum mechanics, as well as educators seeking to deepen their understanding of wave functions and potential wells.

siifuthun
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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 Schrödinger 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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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:
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.
 

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