Eigenvalues and Eigenvectors of a Hermitian operator

[andre220]

Homework Statement

Find the eigenvalues and normalized eigenfuctions of the following Hermitian operator \hat{F}=\alpha\hat{p}+\beta\hat{x}

Homework Equations

In general: $\hat{Q}\psi_i = q_i\psi_i$

The Attempt at a Solution

I'm a little confused here, so for example I don't know if this mean something like $\hat{F}|\psi\rangle =f_i|\psi_i\rangle$ or not. And if not then I don't know where to start.

 

[cwasdqwe]

I would begin by thinking how position and momentum operators act on a given function. This is, remembering that \hat{r} \left|\psi_{i}\right\rangle=r \left|\psi_{i}\right\rangle and \hat{p} \left|\psi_{i}\right\rangle=\frac{\hbar}{i}\vec{\nabla}\left|\psi_{i} \right\rangle. (I have assumed here that p and x mean that!)

 

[BvU]

And yes, you are looking for $\hat{F}|\psi\rangle =f_i|\psi_i\rangle$

 

[andre220]

Okay so we have that $$\hat{F} = \alpha\hat{p} + \beta\hat{x} = \alpha i\hbar\frac{\partial}{\partial x} + \beta\hat{x}.$$ So then $$\hat{F}|\psi\rangle = \left(\alpha i\hbar\frac{\partial}{\partial x} +\beta\hat{x}\right)|\psi\rangle = \alpha i\hbar\frac{\partial}{\partial x}|\psi\rangle + \beta\hat{x}|\psi\rangle.$$ But from here I am not sure where to go.

 

[BvU]

$\hat{F}|\psi\rangle =f_i|\psi_i\rangle$ is where you go. In plain language you want to find functions f (or $\psi$) that satisfy .. f' + ..xf + ..f = 0

 

[cwasdqwe]

$$\hat{F} = \alpha\hat{p} + \beta\hat{x} = \alpha i\hbar\frac{\partial}{\partial x} + \beta\hat{x}.$$ So then $$\hat{F}|\psi\rangle = \left(\alpha i\hbar\frac{\partial}{\partial x} +\beta\hat{x}\right)|\psi\rangle $$

Now maybe the application of what you have in parenthesis on a particular wave function leads to something of the form $$\hat{F}|\psi\rangle = f_{i}|\psi\rangle $$
since that application leads to an integer, in this case our eigenvalue.
So if $$|\psi\rangle=Ae^{ikx}$$ is our wave function written in complex form, where k=\frac{2\pi}{\lambda} is the wave vector, then

$$\hat{F}|\psi\rangle = \left(\alpha i^2 k\hbar +\beta x\right)|\psi\rangle = \left(-\alpha k\hbar +\beta x\right)|\psi\rangle = f|\psi\rangle $$

where $$\alpha, \beta \in \Re $$.

My QM are not very 'fresh', so let's wait for others to help us out ;)

 

[andre220]

Right so I am not seeing in what way the operator(s) can act on an arbitrary function $\psi$. For example if we look at just the momentum term we would get something like $$\alpha\,i\hbar\frac{\partial\psi}{\partial x} = \alpha\,i\hbar\,\psi'.$$

 

[cwasdqwe]

Let's put the terms above straight (in plain text for short)... In the expression F|ψ⟩=ƒ|ψ⟩ we say that |ψ⟩ os an eigenfunction of the operator F, with eigenvalue ƒ.
A normalised function is one such that the integral of the squared function equals one. To do the integral we must take care of which range we are summing, this is, in which range |ψ⟩ is defined (in our case is (-∞, +∞)).

With some format,

\int|\psi|^2dx=\int\psi^{*} \psi dx

where * represents the conjugated (we change sign to the complex unit, i). Let me post this and I meditate the rest of the answer more :D

 

[andre220]

Yes I get that but what I don't see is how anything like your example can be constructed given the fact that the $\psi$ in question has no explicit form. In other words, as far as I can tell $\hat{x}|\psi\rangle = x|\psi\rangle.$ Basically, I don't see how it can be written in any explicit form, if that makes sense.

 

[BvU]

The form of $\psi$ is exactly what you are going to find out, from the requirement that it is an eigenfunction of the given operator. Let's rename $\psi$ to, for instance y, which is a function of x: y(x).

The given operator $\alpha i\hbar\frac{\partial}{\partial x} + \beta\hat{x}$
rewrites to $\alpha i\hbar\frac{dy(x)}{dx} + \beta x\, y(x)$
and the eigenfunction requirement to $\alpha i\hbar\frac{dy(x)}{dx} + \beta x\, y(x) = \lambda\, y(x)\$ with $\lambda$ a (complex) number.

This way it looks more like the differential equation it actually is (and what I "hinted" at in post #5 with .. f' + ..xf + ..f = 0)

 

[andre220]

Ahh okay this makes much more sense now, and thus $\lambda$ is the eigenvalue(s) and $y(x)$ the eigenfunction(s). Thank you. That helps a lot.

 

[cwasdqwe]

BvU's presented it überclear :D Don't forget the normalisation thing, ∫y(x)^2 dx=1, right?

 

[andre220]

Okay sorry, I've been working on this problem (and others) and I think I have the solution, but I am not quite sure. Here is what I have

$$\alpha i\hbar\frac{\partial y(x)}{\partial x} +\beta x y(x)= \lambda y(x)$$
$$\alpha i\hbar\frac{d y(x)}{dx} +(\beta x -\lambda)y(x) = 0$$
Which has the solution:
$$y(x) = A\, exp(i \frac{\beta x^2-2\lambda x}{2\alpha\hbar})$$
So then I think my next step would be (other than normalizing $y(x)$) to take the exponent and solve for $\lambda$ by setting it equal to 0 which would the eigenvalues, however, that doesn't seem correct. What am I missing here?