# Operators in Quantum mechanics: can one swap \Psi and \Psi^*

Tags:
1. Sep 14, 2015

### ManueldelaVaca

1. The problem statement, all variables and given/known data
The demonstration for the momentum operator in Quantum Mechanics goes something like this

<v>=\frac{d}{dt}<x>=\frac{d}{dt} \int x \Psi^* \Psi dx

and then one ends up with

<p>=m<v>=\int \Psi^* (-i \hbar \frac{d}{dx}) Psi dx

however, if you swap the congugates you get
<p>=m<v>=\int \Psi (i \hbar \frac{d}{dx}) Psi^* dx

Can somebody confirm that this is true? When one works with operators the order of the \Psi and \Psi* is important?

Thanks!

2. Relevant equations

3. The attempt at a solution

#### Attached Files:

File size:
5.2 KB
Views:
41
2. Sep 14, 2015

### blue_leaf77

$$-i\hbar \int_{-\infty}^{\infty} \psi^* \frac{\partial \psi}{\partial x} dx = -i \hbar \left(|\psi|^2 \big|_{-\infty}^{\infty} - \int_{-\infty}^{\infty} \psi \frac{\partial \psi^*}{\partial x} dx \right)$$
$$= i\hbar \int_{-\infty}^{\infty} \psi \frac{\partial \psi^*}{\partial x} dx$$
where we have assumed that $\psi$ is square integrable. This is no surprise since momentum is supposed to be Hermitian.

3. Sep 14, 2015

### micromass

Staff Emeritus
But why does $\psi$ vanish at $+\infty$ and $-\infty$? Surely not every square-integrable function has that property?

4. Sep 15, 2015

### blue_leaf77

What might be an example of a square-integrable function which does not vanish at infinities?

5. Sep 15, 2015

### micromass

Staff Emeritus
$$f(x) = x^2 \text{exp}(-x^8\sin^2(x))$$

6. Sep 15, 2015

### blue_leaf77

Why does this function not vanish for x approaches infinity, I think by simple inspection it should? What am I missing here?

EDIT: After seeing the post in http://math.stackexchange.com/quest...ction-tend-to-0-as-its-argument-tends-to-infi, I come to agree with you. It's the infinitely zero width but finite height that may make a function square-integrable although it does not necessarily vanish at infinities. Thanks anyway.

Last edited: Sep 15, 2015
7. Sep 15, 2015

### micromass

Staff Emeritus
If $x=\pi n$ for some integer $n$, then $\sin^2(x) = 0$. Hence for these $x$, we have $f(x) = x^2=\pi^2 n^2$. This does not converge to $0$ if $n$ gets large. In fact, it becomes unbounded.

Last edited: Sep 15, 2015
8. Sep 15, 2015

### micromass

Staff Emeritus
9. Sep 15, 2015

### blue_leaf77

CORRECTION to post #2:
The requirement "where we have assumed that $\psi$ is square integrable" should be narrowed down to $\psi$ which goes to zero when $|x|$ goes to infinity.

10. Sep 16, 2015

### ManueldelaVaca

Thank you all for your useful answers.

$$\Psi$$