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Homework Help: Eigenvalues and Eigenvectors of a Hermitian operator

  1. Sep 17, 2014 #1
    1. The problem statement, all variables and given/known data

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

    2. Relevant equations

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

    3. 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.
  2. jcsd
  3. Sep 18, 2014 #2
    I would begin by thinking how position and momentum operators act on a given function. This is, remembering that [itex]\hat{r} \left|\psi_{i}\right\rangle=r \left|\psi_{i}\right\rangle[/itex] and [itex]\hat{p} \left|\psi_{i}\right\rangle=\frac{\hbar}{i}\vec{\nabla}\left|\psi_{i} \right\rangle[/itex]. (I have assumed here that p and x mean that!)
    Last edited: Sep 18, 2014
  4. Sep 18, 2014 #3


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    And yes, you are looking for ##

    \hat{F}|\psi\rangle =f_i|\psi_i\rangle
  5. Sep 18, 2014 #4
    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.
  6. Sep 18, 2014 #5


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    \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
  7. Sep 18, 2014 #6
    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 [itex]k=\frac{2\pi}{\lambda}[/itex] 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 ;)
  8. Sep 18, 2014 #7
    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 $$\alpha\,i\hbar\frac{\partial\psi}{\partial x} = \alpha\,i\hbar\,\psi'.$$
  9. Sep 18, 2014 #8
    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,

    [itex]\int|\psi|^2dx=\int\psi^{*} \psi dx[/itex]

    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
  10. Sep 18, 2014 #9
    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.
  11. Sep 18, 2014 #10


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    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)
  12. Sep 18, 2014 #11
    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.
  13. Sep 18, 2014 #12
    BvU's presented it überclear :D Don't forget the normalisation thing, ∫y(x)^2 dx=1, right?
  14. Sep 19, 2014 #13
    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?
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