# Homework Help: Commutator maths problem

1. Jan 14, 2010

### 8614smith

1. The problem statement, all variables and given/known data
Determine $$\left[\hat{x},\hat{H}\right]$$

2. Relevant equations

3. The attempt at a solution
$$=x\left(-\frac{\hbar^2}{2m}\frac{\delta^2}{\delta{x^2}}+V\right)\Psi-\left(-\frac{\hbar^2}{2m}\frac{\delta^2}{\delta{x^2}}+V\right)x\psi$$

$$x\left(-\frac{\hbar^2}{2m}\frac{{\delta^2}\psi}{\delta{x^2}}+V\psi\right)+\frac{\hbar^2}{2m}\frac{{\delta^2}(x\psi)}{\delta{x^2}}-Vx\psi$$

$$x\left(-\frac{\hbar^2}{2m}\frac{{\delta^2}}{\delta{x^2}}\right)+\frac{\hbar^2}{2m}\frac{\delta}{\delta{x}}\left(\psi+x\frac{\delta\psi}{\delta{x}}\right)$$

How do you get from the 2nd line to the 3rd? is it an identity of partial differentials? can someone explain how to split the d squared term into two separate parts?

2. Jan 14, 2010

### tiny-tim

Hi 8614smith!

(use "\partial" not "\delta" )

Hint: ∂2/∂x2(xψ) = (∂2x/∂x2)ψ + 2(∂x/∂x)(∂ψ/∂x) + x∂2/∂x2(ψ)

3. Jan 14, 2010

### Altabeh

And one more hint: (∂2x/∂x2)ψ = 0.

4. Jan 14, 2010

### 8614smith

why does this = 0? i would have thought it would equal $$\psi$$

assuming the above quote is correct and the 1st term = 0, I have got to the 3rd line - except that i now have an extra 2 in the equation, how do i get rid of this?

and has your identity worked the same way as expanding expanding ( )^2?

5. Jan 14, 2010

### Altabeh

But I got the right answer. Let's make it look better: $$\frac{\hbar^2}{2m}\frac{{\partial^2} (x\psi)}{\partial{x^2}}=\frac{\hbar^2}{2m}\frac{{\partial}}{\partial{x}}\left(\frac{{\partial} (x\psi)}{\partial{x}}\right)$$. Now use the simple rule (xy)'=x'y+y'x and see what happens.
AB

6. Jan 14, 2010

### Altabeh

It would be like that if it were (∂x/∂x)ψ.

7. Jan 14, 2010

### 8614smith

if i use the product rule on both sides then i just get the same thing on both sides. meaning i just have 0? or have i done that wrong?

The answer quoted for this problem is $$\frac{\hbar^2}{m}\frac{\partial}{\partial{x}}$$

8. Jan 14, 2010

### Altabeh

It means that you have done it right. Let's take the final step:

$$\frac{\hbar^2}{2m}\frac{{\partial^2} (x\psi)}{\partial{x^2}}= \frac{\hbar^2}{2m}\frac{{\partial}}{\partial{x}}\left(\frac{{\partial} (x\psi)}{\partial{x}}\right)= \frac{\hbar^2}{2m}\frac{{\partial}}{\partial{x}}\left(\frac{{\partial} (x)}{\partial{x}}\psi+x\frac{{\partial} (\psi)}{\partial{x}}\right)= \frac{\hbar^2}{2m}\frac{{\partial}}{\partial{x}}\left(\psi+x\frac{{\partial} (\psi)}{\partial{x}}\right)$$. So we ended up getting the expression in the parenthesis of desired answer. You see it, right?!

AB