Proving an equation is an eigenstate of the momentum operator.

[craig.16]

Homework Statement

A free particle (de Broglie wave) may be represented by the wave-function

$\psi$(x)=Ae^ikx​

Show that this is an eigenstate of the momentum operator $\hat{p}$=-$\hbar$$\frac{\delta}{\deltax}$

Homework Equations

$\hat{p}$u_n(x)=a_nu_n(x)
a_n is the eigenvalue
u_n(x) is the corresponding eigenfunction

The Attempt at a Solution

Ok so first I have

$\hat{p}$=-$\hbar$$\frac{\delta}{\deltax}$​

which has the boundary condition that u_n is periodic in the range L. Then the eigenvalue equation is

-i$\hbar\frac{\delta}{\deltax}$$\psi$(x)=a_n$\psi$(x)​

where $\psi$(x) is the eigenfunction u_n(x)
The solution of the equation is then

$\psi$(x)=$\hbar$e^ia_nx
=$\psi$(x)=Ae^ikx​

where k is the eigenvalue a_n and A is $\hbar$ (Plancks Constant)

Ok so what I would like to know is:
Is the above correct?
If so why is it correct?
If its wrong why is it wrong?
What exactly is an eigenfunction and eigenvalue?
What is this range L?

As you can probably guess from the above questions I know absolutely nothing on this topic. I have a sort of understanding on what an eigenfunction and eigenvalue are but after going through revision notes on this stuff numerous times I still can't get my head round the mathematics of it, how everything changes to the way it does. The only reason why I have the above working out is because the revision notes on the topic is very similar to the question but is quite vague on how one thing transitions to the next.

 

[grzz]

When an operator acts on a function, the function may change. But when an operator acts on one of the eigenfunctions of this same operator, the function does not change

 

[craig.16]

So are you saying that -iℏ(δ/deltax) shouldn't change to ℏe^ia_nx since its acting on the eigenfunction of the same operator? Sorry if this is wrong just trying to understand what you said.

 

[grzz]

When an operator acts on one of its own eigenfuctions, it only multiplies the given fuction simply by a number but it does not result in a different function.

 

[vela]

You have$$\hat{p}=-i\hbar\frac{\partial}{\partial x}$$and $\psi(x) = Ae^{ikx}$, so you can show that$$\hat{p}\psi(x) = -i\hbar\frac{\partial}{\partial x}(Ae^{ikx}) = \hbar k \,\psi(x)$$I'll leave it to you to fill in the steps. Applying $\hat{p}$ to this wave function results in a multiple of the wave function, so this wave function is called an eigenfunction or eigenstate of $\hat{p}$, and the multiple, $\hbar k$, is called the eigenvalue.

If you want to go with the approach you originally took, you have$$-i\hbar\frac{d\psi}{dx} = a_n \psi$$
(Since $\psi(x)$ only depends on x, we can replace the partial derivative by a regular derivative.) That equation is separable, so you get
$$-i\hbar\frac{d\psi}{\psi} = a_n\,dx$$
Now you'd want to integrate both sides and solve for $\psi$. The constant of integration eventually turns into A. By comparing your solution to the form of $\psi$ given, you can determine what the eigenvalue $a_n$ is equal to in terms of k.

 

[craig.16]

So going from my original approach if i continue from:

-i$\hbar$$\frac{\delta\psi}{\psi}$=a_ndx

then integrate through I get:

-i$\hbar$ln$\psi$+constant=a_nx+constant

divide through by -i$\hbar$ and move the constant from the LHS to the RHS giving:

ln$\psi$=[itex]\frac{a_nx}{-i\hbar}[/itex]+constant

convert into exponential form:

$\psi$=Ae^{[itex]\frac{a_nx}{-i\hbar}[/itex]}

where k=[itex]\frac{a_n}{\hbar}[/itex]

Is this correct?

Also I'm not sure if this happens on any of your computers/laptops but on mine some of the equations are not displaying properly. If they aren't showing up right what has made it go like that?

 

[vela]

Yes, that looks correct.

Use LaTeX to write the entire expression rather than individual symbols. Don't use sub and sup tags inside itex tags. Use _ for subscripts and ^ for superscripts, e.g. a_n and e^{ikx}.

 

[craig.16]

Ok thanks for that, the equation issue was bugging me for ages haha. Also thanks for the help vela and grzz I think I understand how operators work now.