- #1

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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

- Thread starter naggy
- Start date

- #1

- 60

- 0

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

- #2

- 525

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- #3

- 60

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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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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.

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