Differential Equations - Hermite Polynomials

[mathmannn]

Homework Statement

Here is the entire problem set, but (obviously) you don't have to do it all, if you could just give me a few hints on where to even start, because I am completely lost.

Recall that we found the solutions of the Schrödinger equations
(x^2 - \partial_x ^2) V_n(x) = (2n+1) V_n(x) \quad \quad n = 0,1,2

V_0 (x) = e^{\frac{x^2}{2}} \quad \quad V_n (x) = (x - \partial x)^n V_0(x)

1. Show that for each n there is a polynomial P_n such that V_n(x)=P_n(x)V_0(x)

2. Show that for any smooth function v(x)

(x - \partial_x) v(x) = -e^{\frac{x^2}{2}} \frac{d}{dx} \left ( e^{\frac{-x^2}{2}} v(x) \right )

Homework Equations

During class we went over the commutator of [B,C] and a lot its it's identities.

Also (x^2 + \partial_x ^2) V_n = (2n+1) V_n \rightarrow V_n := (A^*)V_0 \neq 0

The Attempt at a Solution

I have no idea where to even start on this. None of this is in the book and our professor said that each of the problems can be solved using a few "tricks" and then after that they aren't that hard.

 

[susskind_leon]

I think the problem set is really not that "deep" in the sense that you need to have a very good unserstanding of differential equations, Hermite Polynomials or anything else. You just need to know how to do derivatives. Take a good look at rules of differential calculus. Compute the first derivative of V0(x). What do you get? Compute the second derivative. Do you see a pattern? You just need to have a real good look at what you are doing. You could also start with exercise 2 since it's more general and then do exercise 1 after that, using the result of ex. 2.

 

[jackmell]

You could start at 2:

(x-\partial x)^2 V_0(x)=\left(x^2-2x\partial x+\partial^2 x\right) V_0(x)

Keep in mind that powers of operators are higher-order operators and not operators raised to a power right.

How about 3? Then use the Binomial Theorem for n.

For (2), can't you just use the chain-rule and work through that and show that? Remember when you have expressions like:

(x+\partial x) f(x)

that means the operator (x+\partial x) operating on the function x. You know that right? Same dif with the first one raised to higher powers

 

[mathmannn]

Thank you all very much! I got it, I was just thinking too much into the problem!

 

[jackmell]

You could start at 2:

(x-\partial x)^2 V_0(x)=\left(x^2-2x\partial x+\partial^2 x\right) V_0(x)

Keep in mind that powers of operators are higher-order operators and not operators raised to a power right.

Think I'm wrong on that. Rathers it's the operator (x-\partial x) operating on the operator (x-\partial x) so:

(x-\partial x)^2=(x-\partial x)(x-\partial x)=x^2-x\partial x-\partial x (x)+\partial x \partial x

=x^2-x\partial x-1+\partial^2 x

and it's operating on the function f(x), not x as I stated above.

Still gives the polynomial though.