Concerning the momentum operator

  • Thread starter naggy
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  • #1
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If [tex]U[/tex] is an operator so [tex]U\Psi(x)[/tex] = [tex]\Psi(x-a)[/tex].

How can I show that [tex]exp(-iaP/h) = U[/tex]

where P is the momentum operator [tex]P = -ih(d/dx)[/tex]

I sense Fourier analysis
 

Answers and Replies

  • #2
525
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It's matter of Taylor expanding the right hand side. By definition this is just a sum of Psi and its derivatives. You can pull the Psi out of this Taylor series, and what's left is an operator (which is a sum of derivatives) acting on Psi(x). You'll recognise this operator as the series expansion of the exponent U.
 
  • #3
60
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It's matter of Taylor expanding the right hand side. By definition this is just a sum of Psi and its derivatives. You can pull the Psi out of this Taylor series, and what's left is an operator (which is a sum of derivatives) acting on Psi(x). You'll recognise this operator as the series expansion of the exponent U.
What I do know that if I have a function F of an operator then

[tex]F(P)\psi[/tex] = [tex]$\sum_{i} c_iF(\lambda_i)\psi_i[/tex]

where [tex]\lambda_i[/tex] are the eigenvalues of [tex]P[/tex]

and [tex]c_i = <\psi_i,\psi>[/tex]

can I somehow relate all of this to the operator [tex]U[/tex]
 
  • #4
675
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naggy, there might be some connection with your last post.

But what xepma is pointing out is that

[tex]U=exp(-iaP/h)=exp[-a(d/dx)]=\sum{\frac{(-a)^n}{n!}\frac{d^n}{dx^n}}[/tex].

Applying U to [tex]\Psi(x)[/tex] is identical to the Taylor expansion of [tex]\Psi(x-a)[/tex] around x.

That's all there is to it.
 
  • #5
1,838
7
A problem with this proof is that the operator identity is valid on L^2, and thus remains valid when applied to non-analytic psi(x) for which the Taylor expansion around x does not converge to psi(x-a). So, you have to fix this (which is not difficult).
 

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