Mathematical methods of physics problem

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Homework Help Overview

The problem involves verifying an operator identity related to differential operators and exponential functions, specifically in the context of mathematical methods in physics.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss applying both sides of the operator identity to a function and comparing results. There are mentions of using the integrating factor method and exploring the implications of the exponential factors involved.

Discussion Status

Multiple interpretations of the operator identity are being explored, with participants questioning the correctness of the exponential factors and their effects when applied to functions. Some guidance has been offered regarding the application of the operators and the resulting expressions.

Contextual Notes

There are references to the relationship of the problem to Hermite polynomials and operators of creation and annihilation, indicating a broader context in quantum mechanics or mathematical physics.

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Homework Statement



Here's the problem. Verify the operator identity
x - d/dx = -exp (-x^2 / 2) d /dx exp (-x^2 / 2)
Can someone please help me? Thanks :)

Homework Equations





The Attempt at a Solution


 
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I would apply both sides of the equation to a function [itex]f(x)[/itex] and compare the results.

I sense the integrating factor method...
 
ok.thanks for help :)
 
This seems wrong. You have the same factor exp(-x^2/2) both outside and inside the d/dx. The result (when applied to f(x)) will have a factor exp(-x^2) on the right, but not on the left. I think one of the factors on the right should be exp(+x^2/2), so that the two "exp's" finally cancel.

RGV
 
As I said Ray, apply them both to a function f to find:
[tex] xf-\frac{df}{dx}=g(x),\quad e^{-x^{2}/2}\frac{df}{dx} \left( e^{-x^{2}/2}f\right) =g(x)[/tex]
Both are g(x) as when applied to f they should give the same thing. Use the integrating factor [itex]e^{-x^{2}/2}[/itex] and treat it as an ODE.
 
thank you both for helping :)
 
I assume that -exp (-x^2 / 2) d /dx exp (-x^2 / 2) is to be interpreted as an operator that applies to a function f(x), giving -exp(-x^2/2)*(d/dx)[exp(-x^2/2)*f(x)] = x*exp(-x^2)*f(x) - exp(-x^2)*df(x)/dx = exp(-x^2)*[x - d/dx] f(x). The exp(-x^2) arises from exp(-x^2/2)*exp(-x^2/2). Where is my error?

On the other hand (as I suggested before), -exp(+x^2/2) d/dx exp(-x^2/2) applied to f(x) does, indeed, give x*f(x) - df(x)/dx, because we get cancellation: exp(x^2/2)*exp(-x^2/2) = 1.

RGV
 
let's check this, take the first equation, the integrating factor for this is [itex]-e^{-x^{2}/2}[/itex] and the equation becomes:
[tex] \frac{d}{dx}\left( -e^{-x^{2}/2}f(x) \right) =-e^{-x^{2}/2}g(x)[/tex]
Then to isolate g(x), multiply by [itex]-e^{x^{2}/2}[/itex] to get:
[tex] -e^{x^{2}/2} \frac{d}{dx}\left( -e^{-x^{2}/2}f(x) \right) =g(x)[/tex]
So there is a difference in signs as Ray said.
 
The problem is related to the Hermite polynomials and operators of creation and annihilation. The creation operator:
[itex]a^{\dagger}\equiv \frac{1}{\sqrt{2}}\left(x- \frac{d}{dx}\right)[/itex]
I played with the Rodrigues formula for Hermite poyinomials (for n=1?):
[itex]H_n (x)=(-1)^n e^{x^2} \frac{d^n}{dx^n}e^{-x^2}[/itex]
and
[itex]\psi_n(x)=\frac{1}{\sqrt{2^n n!\sqrt{\pi}}}e^{-\frac{x^2}{2}}H_n(x)[/itex]
but run into some problems.
 
Last edited:

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