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Problem from Intro to Quantum Mechanics by Griffiths (I know near nothing)

  • Thread starter WraithM
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Okay, here's a quantum mechanics problem. I am just starting with quantum, so I have no idea about this stuff. I am requesting guidence or a push in the right direction, not nessicarily a complete answer, please. Also, I don't know if this fits into the Advanced Physics help forum. I've never posted on the homework help forum before. If this is too simple, please let me know.

Homework Statement



A particle of mass m is in the state

[tex]\Psi (x, t) = A e^{-a[(m x^2/\hbar) + i t]}[/tex],

where A and a are positive real constants.

(a) Find A.


Homework Equations



Shroedinger's equation?

[tex]\int_{-\infty}^{+\infty} \mid \psi \mid ^2 dx = 1[/tex]

Maybe?

The Attempt at a Solution



I guess my plan was to take the complex conjugate of the wave function there, multiply them, and then intregrate with [tex]\int_{-\infty}^{+\infty} \mid \psi \mid ^2 dx = 1[/tex] to get the answer for A. I ran into troubles along the way, and I have some theories as to where I messed up.

Firstly, I think I took the complex conjugate incorrectly. My knowledge of complex numbers is extremely rusty :( I haven't really studied that since perhaps my sophomore year in high school...

Here's what I said it was:

[tex]\Psi^* = A e^{-a[(m x^2/\hbar) - i t]}[/tex]

So, assuming I did that correctly (which is not a very good assumption :frown:), I moved foreward.

[tex]\mid \psi \mid ^2 = A^2 e^{-2 a m x^2/\hbar}[/tex]

Okay, and that's a gaussian, so the integration is tricky. I assume that this is not the proper way of moving foreward because when I tried to think of how to do the integration I got confused.

Does anybody have a good idea of how to move foreward with this problem?

-WraithM
 
Last edited:

Answers and Replies

  • #2
dextercioby
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Can you integrate a gaussian ? If not, then use a table of integrals. I'm sure you can find one in every QM book.
 
  • #3
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The integral is in the answer of an "error function" which I frankly have no idea how to deal with, but I like your idea of a table of integrals. I have a really solid table of integrals with 728 of them :D I guess I'll try that.

If anybody knows how to do the integral

[tex]\int_{-\infty}^{+\infty} A^2 e^{-2 a m x^2/\hbar} dx[/tex]

it would help me very much.
 
  • #4
dextercioby
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There's no 'error' function answer to this one, because the limits are + and - infinity. The solving of the integral

[tex] \int_{-\infty} ^{+\infty} e^{-x^2} {} dx [/tex]

is one of the oldest tricks in any analysis book. However, I'm sure you can find its answer in your QM book.
 
  • #5
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The integral is in the answer of an "error function" which I frankly have no idea how to deal with, but I like your idea of a table of integrals. I have a really solid table of integrals with 728 of them :D I guess I'll try that.

If anybody knows how to do the integral

[tex]\int_{-\infty}^{+\infty} A^2 e^{-2 a m x^2/\hbar} dx[/tex]

it would help me very much.
use substitution and then:

http://en.wikipedia.org/wiki/Gaussian_integral#Brief_proof
 
  • #6
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Win!

[tex]\int_{0}^{\infty} e^{-b^2 x^2} dx = \frac{\sqrt{\pi}}{2b}[/tex]

My 728 best friends never fail.

[tex]\int_{-\infty}^{+\infty} A^2 e^{-2 a m x^2/\hbar} dx = 2 \int_{0}^{\infty} A^2 e^{-2 a m x^2/\hbar} dx[/tex]

So,

[tex]b^2 = 2 a m/\hbar[/tex]

And

[tex]2 \int_{0}^{\infty} A^2 e^{-2 a m x^2/\hbar} dx = A^2 \sqrt{\frac{\pi \hbar}{2 a m}} = 1[/tex]

Therefore,

[tex]A = (\frac{2 a m}{\pi \hbar})^{1/4}[/tex]

My math is rusty :( I'm not going to do well figuring out this quantum mechanics, lol.

Thank you very much for the help!
 

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