QM Vector Spaces and States

Homework Statement

The vector |p> is given by the function x+2x2 and the operator A = 1/x * d/dx, with x = [0,1].

a) Compute the norm of |p>

b) Compute A|p>. Does A|p> belong to the VS of all real valued, continuous functions on the interval x = [0,1]?

c) Find the eigenvalues and eigenvectors of A.

-

The Attempt at a Solution

a) For the norm, I have that it should be $$\sqrt{<p|p>}$$ but I don't know how to find the scalar product of x+2x2 with itself

b) A|p> is just 1/x + 4, which isn't continuous at x=0 so no?

c) I have no idea how to turn the operator into a matrix. Once I have the matrix, it should be easy but what is the matrix form of A = 1/x * d/dx? Or do I need to use the eigenvalue equation and say that A|p> = eigenvalue|p>?

Any help would be appreciated.
Thanks!

Last edited:

Answers and Replies

dextercioby
Science Advisor
Homework Helper
I sense somehow that the Hilbert space is $L^2 \left(\left[0,1\right], dx\right)$, so computing the norm of that vector should be easy.

For the point c), you don't need any matrix, just the eigenvalue equation.

I sense somehow that the Hilbert space is $L^2 \left(\left[0,1\right], dx\right)$, so computing the norm of that vector should be easy.

For the point c), you don't need any matrix, just the eigenvalue equation.

Ah so for the norm, it's just $$\bigg(\int_0^1 (x+2x^2)^2 \mathrm{d}x\bigg)^\frac{1}{2}?$$

The eigenvalue equation will be $$\frac{1}{x}\frac{\mathrm{d}}{\mathrm{d}x} |\psi> = \lambda |\psi>$$ but then how do you solve this, or do you assume that the eigenvector is |p>?

(Actually, |p> clearly isn't an eigenvector...)

Last edited:
Oh wait, do you just solve it like a 1st order ODE to get $$|\psi> = Aexp\bigg(\frac{x^2}{2}\bigg)$$ which means that the eigenvalue is 1?

Last edited:
dextercioby
Science Advisor
Homework Helper
Yes to the first post, kind of yes to to the second post when it comes to the wavefunction (actually $\lambda$ should also be in the exponential), however, the spectrum of the operator (possible values of $\lambda$) is the entire complex plane.

Last edited:
Yes to the first post, kind of yes to to the second post when it comes to the wavefunction (actually $\lambda$ should also be in the exponential), however, the spectrum of the operator (possible values of $\lambda$) is the entire complex plane.

Oh there is a $\lambda$, I took it out because it looked a bit strange. So $\lambda$ can be anything?

dextercioby
Science Advisor
Homework Helper
Yes.

Yes.

Hmm interesting. Thanks a lot for your help!