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

[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 anybody offer some methods to handle this question? Thank you very much.

 

[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).

 

[ee_mike]

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).

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.

 

[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 ;)

 

[ee_mike]

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 ;)

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