# A question about solving the energy eigenvalue of a given Hamiltonian operator

1. Apr 2, 2009

### ee_mike

The problem is

A particle of mass m and electric charges q can move only in one dimension and is subject to a harmonic force and a homogeneous electrostatic field. The Hamiltonian operator for the system is
H= p2/2m +mw2/2*x2 - qεx

a. solve the energy eigenvalue problem
b. if the system is initially in the ground state of the unperturbed harmonic oscillator, ket= |0>, what is the probability of finding it in the ground state of the full Hamiltonian?

Could any body offer some methods to handle this question? Thank you very much.

2. Apr 2, 2009

### xepma

I take it you have solved the harmonic oscillator.

This question is pretty much the same. The trick is to complete the square of the coordinate x-dependent part. That is, turn this:

$$ax^2 + bx$$

into this:

$$a(x+\frac{b}{2a})^2 - (\frac{b}{2a})^2$$

From there you should recognise the harmonic oscillator, only this time with some shifted origin plus a constant term in you Hamiltonian (which you shouldn't throw away).

3. Apr 2, 2009

### ee_mike

Could you explain more detailedly? Because I wonder what is the eigenvalue of $$a(x+\frac{b}{2a})^2 - (\frac{b}{2a})^2$$

Thank you very much, you really helped me.

4. Apr 2, 2009

### xepma

So you start out with the basic eigenvalue equation:

$$H\psi = E \psi$$

where, for simplicity, we have

$$H = T + ax^2 + bx$$

where T is short for the kinetic energy.
Filling this in gives you the following eigenvalue equation:

$$(T + ax^2 + bx )\psi = E \psi$$

Now you perform the trick I explained earlier, which give you:

$$\left(T + a(x+\frac{b}{2a})^2-(\frac{b}{2a})^2\right)\psi = E \psi$$
or:
$$\left(T + a(x+\frac{b}{2a})^2\right)\psi = (E+(\frac{b}{2a})^2) \psi$$

Compare this with the standard form of a harmonic oscillator:

$$\left(T + \frac{m\omega}{2} y^2\right)\psi' = \epsilon \psi'$$

As you can see, you now have a shifted origin of the harmonic oscillator, and a shifted energy.

I already brought you more than halfway there, so good luck with the rest of the steps ;)

5. Apr 2, 2009

### ee_mike

You are so powerful. I have been doing on this problem since this morning. Thank you very much.