Eigenvalues for a Hamiltonian

  • Thread starter umagongdi
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  • #1
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Homework Statement



Consider the Hamiltonian

[tex]\hat{}H[/tex] = [tex]\hat{}p[/tex]2/2m + (1/2)mω2[tex]\hat{}x[/tex]2 + F[tex]\hat{}x[/tex]

where F is a constant. Find the possible eigenvalues for H. Can you give a physical
interpretation for this Hamiltonian?

Homework Equations



The Attempt at a Solution



I don't think you can put H in matrix form?

Can i use the following?
HΨ=λΨ

HΨ = p2Ψ/2m+(1/2)mω2x2Ψ+FxΨ = λΨ?
I think i am using the wrong method.

What does "give a physical interpretation..." mean?

thanks in advance
 
Last edited:

Answers and Replies

  • #2
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Haha Russo still up to his tricks

you have to complete the square in the position operator terms to get a new position operator plus a constant term added onto the hamiltonian.

This added term means that it is a shifted harmonic oscillator
 
  • #3
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Haha Russo still up to his tricks

lol that's awsome that we went/go to the same uni, what do you do now Masters/PhD here?

As for the question, i don't understand how completing the square will help us find the eigenvalues.

Looking at

(1/2)mω2X2+ FX2

We can factorise the (1/2)mω2X2

(1/2)mω2X2(X2+FX/(1/2)mω2X2)

Looking at:

(X2+FX/(1/2)mω2X2)=(X+F/mω)2)-(F/mω2)2)

Subbing everything into original equation we get:

H=P2/2m +(1/2)mω2(X+F/mω2)2-(1/2)(F2/mω2)
 
  • #4
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Can we look at HΨ=?Ψ

We can rearrange and solve for Ψ?
 
  • #5
vela
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Two things:

1. If you add a constant to the Hamiltonian, what does that do to the eigenfunctions and eigenvalues?

2. Initially you had the potential V(x) and now it looks like V(x+a), where a is a constant. Physically, what does x going to x+a represent? What effect should that have on the solutions?
 
  • #6
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I'm at Cambridge doing PartIII and missing the simple life at Queen Mary

anyway the Harmonic oscillator can be rearanged in terms of the number operator

[tex] \hat{H} = \frac {\hat{p}^2}{2m} +\frac {1}{2} m \omega^2 \hat{x}^2 = \hbar \omega (a^{\dagger} a +\frac {1}{2}) = \hbar \omega(\hat{n} + \frac{1}{2}) [/tex]

adding a constant to the hamiltonian does not change the eiegenstates only the eigenvalues
 
  • #7
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Two things:

1. If you add a constant to the Hamiltonian, what does that do to the eigenfunctions and eigenvalues?

adding a constant to the hamiltonian does not change the eigenstates only the eigenvalues

In this case is the eigenfunction HΨ=EΨ or eigenstate, what is the difference, are they the same? I am a bit confused is the whole equation HΨ=EΨ called a eigenfunction or a specific part of it?

I think i finally understood how to do it, i used the ladder operator method. I obtained this part

[tex] \hat{H} = \frac {\hat{p}^2}{2m} +\frac {1}{2} m \omega^2 \hat{x}^2 = \hbar \omega (a^{\dagger} a +\frac {1}{2})

I dont know how to do the following:

[tex]\hbar \omega (a^{\dagger} a +\frac {1}{2}) = \hbar \omega(\hat{n} + \frac{1}{2}) [/tex]

Finally starting to understand QM, thanks you guys.
 
Last edited:
  • #8
213
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the equation [tex] \hat{H} \Psi = E \Psi [/tex] is used to define the eigenfinction [tex] \Psi [/tex] with eigenvalues E

as for the number operator we derive that from the fact that

[tex] a \left| n \right \rangle = \sqrt{n} \left| n-1 \right \rangle [/tex]

[tex] a^{\dagger} \left| n \right \rangle = \sqrt{n+1} \left| n+1 \right \rangle [/tex]

these come from the conditions

[tex] a \left| 0 \right \rangle = 0 [/tex]

[tex] a^{\dagger} \left| 0 \right \rangle \propto \left| 1 \right \rangle [/tex]

and from the fact that we want the number operator to reproduce the eigenvalues of the QHO i.e. [tex] \hbar \omega (n + 1/2) [/tex]
 

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