# Eigenvalues for a Hamiltonian

## Homework Statement

Consider the Hamiltonian

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

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

## 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?

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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

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)

Can we look at HΨ=?Ψ

We can rearrange and solve for Ψ?

vela
Staff Emeritus
Homework Helper
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?

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

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

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

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

$$\hat{H} = \frac {\hat{p}^2}{2m} +\frac {1}{2} m \omega^2 \hat{x}^2 = \hbar \omega (a^{\dagger} a +\frac {1}{2}) I dont know how to do the following: [tex]\hbar \omega (a^{\dagger} a +\frac {1}{2}) = \hbar \omega(\hat{n} + \frac{1}{2})$$

Finally starting to understand QM, thanks you guys.

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the equation $$\hat{H} \Psi = E \Psi$$ is used to define the eigenfinction $$\Psi$$ with eigenvalues E

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

$$a \left| n \right \rangle = \sqrt{n} \left| n-1 \right \rangle$$

$$a^{\dagger} \left| n \right \rangle = \sqrt{n+1} \left| n+1 \right \rangle$$

these come from the conditions

$$a \left| 0 \right \rangle = 0$$

$$a^{\dagger} \left| 0 \right \rangle \propto \left| 1 \right \rangle$$

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