# Finding relationships of inner products with operator

1. Mar 19, 2015

### Maylis

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

2. Relevant equations

3. The attempt at a solution
a) I am having some trouble understanding the notation. I'm uncertain whether it should be
$$\langle {f} | \hat {O_{2}} | g \rangle = \int_{-\infty}^{\infty} f^{*}g \frac {dg}{dx} dx$$
or
$$\langle {f} | \hat {O_{2}} | g \rangle = \int_{-\infty}^{\infty} f^{*}x \frac {dg}{dx} dx$$

b) Once again, the notation is really screwing with my head. I know to start,
$$[\hat {O_{1}}, \hat {O_{2}}] \psi = (\hat {O_{1}} \hat {O_{2}} - \hat {O_{2}} \hat {O_{1}}) \psi$$
$$= \hat {O_{1}}(\hat {O_{2}} \psi) - \hat {O_{2}}(\hat {O_{1}} \psi)$$
$$= \hat {O_{1}}(x \frac {d \psi}{dx}) - \hat {O_{2}}(x^{3} \psi)$$

From here, I am unsure how to do this. My thoughts are either
$$= x^{3} \Big (x \frac {d \psi}{dx} \Big ) - x \frac {d(x^{3} \psi)}{dx}$$
or
$$= x^{3} \frac {d^{3} \psi}{dx^{3}} - ???$$
Wouldn't even know what to use for the second term.

I will assume it's the first one since at least I know what to do with that one. Using product rule,
$$= x^{4} \frac {d \psi}{dx} - x \Big [ x^{3} \frac {d \psi}{dx} + \psi \frac {d (x^{3})}{dx} \Big ]$$
$$= x^{4} \frac {d \psi}{dx} - x^{4} \frac {d \psi}{dx} - 3x^{3} \psi$$
Therefore, the commutator is $-3x^{3}$

Last edited: Mar 19, 2015
2. Mar 19, 2015

### Staff: Mentor

This one is correct.

This one is correct.

Correct!

Maybe it would help you seeing $\hat {O_{2}}$ as simply
$$\hat {O_{2}} = s \frac{d}{dx}$$
and when you "multiply" from the left with $d/dx$, it means you take the derivative of the function that is to the right of the operator.

3. Mar 20, 2015

### Maylis

Okay then, now I am having some trouble integrating this

Starting from

$$\langle {f} | \hat {O_{2}} | g \rangle = \int_{-\infty}^{\infty} f^{*}x \frac {dg}{dx} dx$$
I would probably try to use integration by parts. Choosing $u = f^{*}x$, $dv = \frac {dg}{dx} dx$, $du = \frac {d(f^{*}x)}{dx}$, $v = g$
$$uv - \int v \hspace {0.03 in} du$$
$$= f^{*}xg - \int_{-\infty}^{\infty} g(f^{*} + x \frac {df^{*}}{dx})$$
$$= f^{*}xg \bigg |_{-\infty}^{\infty} - \int_{-\infty}^{\infty} gf^{*}dx - \int_{-\infty}^{\infty}g x \frac {df^{*}}{dx}$$

Last edited: Mar 20, 2015
4. Mar 20, 2015

### Staff: Mentor

You simply forgot it when calculating $du$:
$du = \frac {d(f^{*}x)}{dx} dx$

5. Mar 20, 2015

### Maylis

Okay, I caught that error. So now I got

$$= f^{*}xg \bigg |_{-\infty}^{\infty} - gf^{*}x \bigg |_{-\infty}^{\infty} - \int_{-\infty}^{\infty}g x \frac {df^{*}}{dx} dx$$
The first two terms cancel, and I am not sure how to evaluate the third term. Also, I am not sure how to manipulate this third integral to show whether or not $\hat {O_{2}}$ is hermitian or not. would it just be
$$- \int_{-\infty}^{\infty} x \frac {df^{*}}{dx}g \hspace {0.03 in} dx = - \int_{-\infty}^{\infty} (x \frac {df^{*}}{dx})^{*} g \hspace {0.03 in} dx$$

I honestly have no clue how to manipulate to show if it is hermitian or not...for the example of showing if the momentum operator is hermitian or not, I know they just switched the sign of the imaginary part and put a conjugate, then said it was hermitian, but I don't understand it.

I calculate $\langle g | \hat {O_{2}} | f \rangle$ up to the same point and reach
$$- \int_{-\infty}^{\infty} x \frac {dg^{*}}{dx}f \hspace {0.03 in} dx$$

I suppose there is something between these two that should determine the relationship between the two and that will answer the question of whether or not the operator is hermitian. How can I show that one is the adjoint of the other?

Last edited: Mar 20, 2015
6. Mar 20, 2015

### Staff: Mentor

The way to apply hermitian conjugation to a bracket is
$$\left( \langle f | \hat{O} | g \rangle \right)^\dagger = \langle g | \hat{O}^\dagger | f \rangle$$
Use that to determine if $\hat{O}_2$ is hermitian.

7. Mar 20, 2015

### Staff: Mentor

I just noticed that that second term there is incorrect.