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Quantum mechanics, wavefunction problem

  • Thread starter fluidistic
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  • #1
fluidistic
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Homework Statement


Consider the wavefunction [itex]\Psi (x,t)=c_1 \psi _1 (x)e^{-\frac{iE_1t}{\hbar}}+c_2 \psi _2 (x)e^{-\frac{iE_2t}{\hbar}}[/itex] where [itex]\psi _1 (x)[/itex] and [itex]\psi _2 (x)[/itex] are normalized and orthogonal. Knowing [itex]\Psi (x,0)[/itex], find the values of [itex]c_1[/itex] and [itex]c_2[/itex].


Homework Equations



[itex]C^2 \int _{-\infty}^{\infty} |\Psi (x,t)|^2 dx=1[/itex]. I also know that the product of psi 1 by psi 2 is worth 0 (they are orthogonal) so this simplifies the expression to integrate.
But I'm still left with the integration of both lower case psi functions that I don't know how to handle.

The Attempt at a Solution


I'm thinking on how to use the fact that I know Psi at t=0 but so far I'm out of ideas.
 

Answers and Replies

  • #2
ehild
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[itex]\psi _i (x)[/itex]-s are orthonormal, what does it mean on the integrals of [itex]\psi _i ^* (x) \psi _j (x)[/itex]?

ehild
 
  • #3
fluidistic
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[itex]\psi _i (x)[/itex]-s are orthonormal, what does it mean on the integrals of [itex]\psi _i ^* (x) \psi _j (x)[/itex]?

ehild
Sorry for being almost 1 month late. That they are worth 1?

Edit: Is what I've done right?:
[itex]\int _{-\infty}^{\infty} \Psi _1 (x) \Psi (x,0)dx=c_1 \int _{-\infty}^{\infty} \Psi _1 ^2 (x)dx=c_1[/itex].
In the same fashion I get [itex]c_2=\int _{-\infty}^{\infty} \Psi _2 (x) \Psi (x,0)dx[/itex].
But I didn't use any complex conjugate... I must be missing something.
 
Last edited:
  • #4
vela
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That's almost correct. It should be
[tex]c_i = \int \psi_i^*(x)\Psi(x,0)\,dx[/tex]When you calculate the inner product, you use the complex conjugate of the first function in the integrand.
 
  • #5
fluidistic
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That's almost correct. It should be
[tex]c_i = \int \psi_i^*(x)\Psi(x,0)\,dx[/tex]When you calculate the inner product, you use the complex conjugate of the first function in the integrand.
Let's see if I understand:
[itex]\int _{- \infty}^{\infty} \Psi _1 ^* (x) \Psi (x,0)dx=c_1 \int _{- \infty}^{\infty} \Psi _1 ^* (x) \Psi _1 (x)dx=c_1 \int _{-\infty} ^{\infty} |\Psi _1 (x)|^2 dx=c_1[/itex].
Am I right if I say that since [itex]\Psi _1(x)[/itex] and [itex]\Psi _2 (x)[/itex] are orthogonal, so is [itex]\Psi _1^* (x)[/itex] with [itex]\Psi _2 (x)[/itex]? Because I'm using this fact!
Thanks for your help guys.
 
  • #6
vela
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Yup. When you say the two are orthogonal, you mean that
[tex]\int \psi_1^*(x)\psi_2(x)\,dx = 0.[/tex]It does not mean that
[tex]\int \psi_1(x)\psi_2(x)\,dx = 0.[/tex]The complex conjugation is a vital part of taking the inner product.
 
  • #7
fluidistic
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Yup. When you say the two are orthogonal, you mean that
[tex]\int \psi_1^*(x)\psi_2(x)\,dx = 0.[/tex]It does not mean that
[tex]\int \psi_1(x)\psi_2(x)\,dx = 0.[/tex]The complex conjugation is a vital part of taking the inner product.
Oh, I understand. Wow, thank so you much. My understanding was so poor before I asked help for this problem... now things are getting clearer.
 

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