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QM Vector Spaces and States

  • Thread starter ryanwilk
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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.

Homework Equations



-

The Attempt at a Solution



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

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!
 
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Answers and Replies

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

For the point c), you don't need any matrix, just the eigenvalue equation.
 
  • #3
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I sense somehow that the Hilbert space is [itex] L^2 \left(\left[0,1\right], dx\right) [/itex], 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 [tex]\bigg(\int_0^1 (x+2x^2)^2 \mathrm{d}x\bigg)^\frac{1}{2}?[/tex]

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

(Actually, |p> clearly isn't an eigenvector...)
 
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Oh wait, do you just solve it like a 1st order ODE to get [tex]|\psi> = Aexp\bigg(\frac{x^2}{2}\bigg)[/tex] which means that the eigenvalue is 1?
 
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  • #5
dextercioby
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Yes to the first post, kind of yes to to the second post when it comes to the wavefunction (actually [itex] \lambda [/itex] should also be in the exponential), however, the spectrum of the operator (possible values of [itex] \lambda [/itex]) is the entire complex plane.
 
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  • #6
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Yes to the first post, kind of yes to to the second post when it comes to the wavefunction (actually [itex] \lambda [/itex] should also be in the exponential), however, the spectrum of the operator (possible values of [itex] \lambda [/itex]) is the entire complex plane.
Oh there is a [itex] \lambda [/itex], I took it out because it looked a bit strange. So [itex] \lambda [/itex] can be anything?
 
  • #7
dextercioby
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Yes.
 
  • #8
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Yes.
Hmm interesting. Thanks a lot for your help! :smile:
 

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