Homework Help: Mathematical methods of physics problem

1. Jul 19, 2011

gorved

1. The problem statement, all variables and given/known data

Here's the problem. Verify the operator identity
x - d/dx = -exp (-x^2 / 2) d /dx exp (-x^2 / 2)
2. Relevant equations

3. The attempt at a solution
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Jul 19, 2011

hunt_mat

I would apply both sides of the equation to a function $f(x)$ and compare the results.

I sense the integrating factor method...

3. Jul 19, 2011

gorved

ok.thanks for help :)

4. Jul 19, 2011

Ray Vickson

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

5. Jul 19, 2011

hunt_mat

As I said Ray, apply them both to a function f to find:
$$xf-\frac{df}{dx}=g(x),\quad e^{-x^{2}/2}\frac{df}{dx} \left( e^{-x^{2}/2}f\right) =g(x)$$
Both are g(x) as when applied to f they should give the same thing. Use the integrating factor $e^{-x^{2}/2}$ and treat it as an ODE.

6. Jul 19, 2011

gorved

thank you both for helping :)

7. Jul 20, 2011

Ray Vickson

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

8. Jul 20, 2011

hunt_mat

let's check this, take the first equation, the integrating factor for this is $-e^{-x^{2}/2}$ and the equation becomes:
$$\frac{d}{dx}\left( -e^{-x^{2}/2}f(x) \right) =-e^{-x^{2}/2}g(x)$$
Then to isolate g(x), multiply by $-e^{x^{2}/2}$ to get:
$$-e^{x^{2}/2} \frac{d}{dx}\left( -e^{-x^{2}/2}f(x) \right) =g(x)$$
So there is a difference in signs as Ray said.

9. Jul 20, 2011

Damir1899

The problem is related to the Hermite polynomials and operators of creation and annihilation. The creation operator:
$a^{\dagger}\equiv \frac{1}{\sqrt{2}}\left(x- \frac{d}{dx}\right)$
I played with the Rodrigues formula for Hermite poyinomials (for n=1?):
$H_n (x)=(-1)^n e^{x^2} \frac{d^n}{dx^n}e^{-x^2}$
and
$\psi_n(x)=\frac{1}{\sqrt{2^n n!\sqrt{\pi}}}e^{-\frac{x^2}{2}}H_n(x)$
but run into some problems.

Last edited: Jul 20, 2011