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

[ManueldelaVaca]

Homework Statement

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!

Homework Equations

The Attempt at a Solution

 

[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.

 

[micromass]

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

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

 

[blue_leaf77]

Surely not every square-integrable function has that property?

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

 

[micromass]

[tex]f(x) = x^2 \text{exp}(-x^8\sin^2(x))[/tex]

 

[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.

 

[micromass]

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.

 

[micromass]

Anyway, this paper explains the issue completely http://arxiv.org/pdf/quant-ph/9907069v2.pdf

 

[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.

 

[ManueldelaVaca]

Thank you all for your useful answers.

$$\Psi$$