What Are the Eigenvalues of a Hamiltonian with a Linear Term?

[umagongdi]

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?

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

 

[sgd37]

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

 

[umagongdi]

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)

 

[umagongdi]

Can we look at HΨ=?Ψ

We can rearrange and solve for Ψ?

 

[vela]

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?

 

[sgd37]

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

 

[umagongdi]

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 don't know how to do the following:

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

Finally starting to understand QM, thanks you guys.

 

[sgd37]

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. $$\hbar \omega (n + 1/2)$$