Hamiltonian of a 1D Linear Harmonic Oscillator

• Patrick McBride
In summary: I hadn't heard of the expectation value before now, but I definitely understand now :)In summary, the Hamiltonian for a one-dimensional linear harmonic oscillator can be shown to be expressed in two forms, with the second form being derived from the Schrodinger equation. The momentum and position operators, P and X, respectively, are related to the classical oscillator frequency and can be used to simplify the equations. The Hamiltonian is defined up to an arbitrary constant, which is usually taken to be zero, but in this case was chosen to make the ground state energy zero.
Patrick McBride

Homework Statement

Show that for the one-dimensional linear harmonic oscillator the Hamiltonian is:
[; H = \frac{1}{2}[P^2+\omega ^2 X^2]-\frac{1}{2}\omega \hbar ;]

[; =\frac{1}{2}[P+i\omega X][P-i\omega X]+\frac{1}{2} \omega \hbar ;]

where P, X are the momentum and position operators, respectively, and [; X= \sqrt{m}x ;].

Homework Equations

[; H = \frac{1}{2}[P^2+\omega ^2 X^2]-\frac{1}{2}\omega \hbar ;]

[; =\frac{1}{2}[P+i\omega X][P-i\omega X]+\frac{1}{2} \omega \hbar ;]

[; X= \sqrt{m}x ;] [; P = p/ \sqrt{m}} ;]

Sch. Eqn.: [;-\frac{\hbar}{2m}\frac{d^2\Psi }{dx^2}+\frac{1}{2}m\omega^2x^2\Psi = E\Psi ;].

The Attempt at a Solution

[/B]
So as from what I can tell [; [P+i\omega X] ;] refers to the +x direction and [;
[P-i\omega X] ;] is in the -x direction. The momentum operator is going to be [; P = p/ \sqrt{m}} ;].Sch. Eqn.: [;-\frac{\hbar}{2m}\frac{d^2\Psi }{dx^2}+\frac{1}{2}m\omega^2x^2\Psi = E\Psi ;].

So what I assume is Sch.Eqn. must be solved algebraically. From there we can get our Hamiltonian.

Any suggestions?

Patrick McBride said:

Homework Statement

Show that for the one-dimensional linear harmonic oscillator the Hamiltonian is:
$$H = \frac{1}{2}[P^2+\omega ^2 X^2]-\frac{1}{2}\omega \hbar$$

$$=\frac{1}{2}[P+i\omega X][P-i\omega X]+\frac{1}{2} \omega \hbar$$

where P, X are the momentum and position operators, respectively, and ##X= \sqrt{m}x##.

Homework Equations

$$H = \frac{1}{2}[P^2+\omega ^2 X^2]-\frac{1}{2}\omega \hbar$$

$$=\frac{1}{2}[P+i\omega X][P-i\omega X]+\frac{1}{2} \omega \hbar$$

$$X= \sqrt{m}x ~~~~~P = p/ \sqrt{m}$$

Sch. Eqn.:

$$-\frac{\hbar}{2m}\frac{d^2\Psi }{dx^2}+\frac{1}{2}m\omega^2x^2\Psi = E\Psi$$

The Attempt at a Solution

[/B]
So as from what I can tell ## [P+i\omega X] ## refers to the +x direction and ##
[P-i\omega X] ## is in the -x direction. The momentum operator is going to be ## P = p/ \sqrt{m} ##.Sch. Eqn.: ##-\frac{\hbar}{2m}\frac{d^2\Psi }{dx^2}+\frac{1}{2}m\omega^2x^2\Psi = E\Psi ##.

So what I assume is Sch.Eqn. must be solved algebraically. From there we can get our Hamiltonian.

Any suggestions?

I corrected all your TeX tags so that the equations will show up properly.

First:, the ## P \pm i \omega x ## have nothing to do with momentum along +x or -x, each P is actually ##P_x## and may be positive or negative.

The Hamiltonian is simply the operator H in ## H \psi = E \psi ## so you may read it off directly from the Schrodinger equation that you wrote. Then to show the second form, you may simply expand out the second form and use what you know about the commutator of P and X.

Ok so I had a crack at it and this is what I've gotten:

## \frac{P^2}{2m} +\frac{1}{2}kx^2 =H ## Harmonic Oscillator Hamiltonian

## -\frac{\hbar^2}{2m} \frac{d^2\Psi }{dx^2}+\frac{1}{2}kx^2 \Psi = E \Psi ## Schrodinger Equation for energy eigenstates.

## \omega =\sqrt{\frac{k}{m}} ## Replace spring constant with classical oscillator frequency.

## \frac{p^2}{2m} = \frac{1}{2}(\frac{p}{\sqrt{m}})^2 = \frac{1}{2}P^2 ## Simplifying using the momentum operator ##P##.

## X^2 = (\sqrt{m}x)^2 = mx^2 ## We can do the same for our position operator ##X##.

## H =\frac{1}{2}P^2 + \frac{1}{2}\omega ^2 X^2 = \frac{1}{2}(P^2 +\omega^2 X^2) ## Subbing into our equation for ##H##.

That explains everything except ##-1/2 \omega \hbar ## from the first equation.
Am I missing something?

Patrick McBride said:
Ok so I had a crack at it and this is what I've gotten:

## \frac{P^2}{2m} +\frac{1}{2}kx^2 =H ## Harmonic Oscillator Hamiltonian

## -\frac{\hbar^2}{2m} \frac{d^2\Psi }{dx^2}+\frac{1}{2}kx^2 \Psi = E \Psi ## Schrodinger Equation for energy eigenstates.

## \omega =\sqrt{\frac{k}{m}} ## Replace spring constant with classical oscillator frequency.

## \frac{p^2}{2m} = \frac{1}{2}(\frac{p}{\sqrt{m}})^2 = \frac{1}{2}P^2 ## Simplifying using the momentum operator ##P##.

## X^2 = (\sqrt{m}x)^2 = mx^2 ## We can do the same for our position operator ##X##.

## H =\frac{1}{2}P^2 + \frac{1}{2}\omega ^2 X^2 = \frac{1}{2}(P^2 +\omega^2 X^2) ## Subbing into our equation for ##H##.

That explains everything except ##-1/2 \omega \hbar ## from the first equation.
Am I missing something?
Good work.

The hamiltonian is actually defined up to an additive constant, that is one can *define* the Hamiltonian to be what you got plus any constant energy. This added constant is arbitrary. It is usually taken to be zero, in which case your expression would be the final answer. But what they did here is to choose the constant in such a way that the expectation value ##\langle 0 | H | 0 \rangle =0## , in other words they wanted the ground state energy to be zero (they should have mentioned that in the question, if they did not, they were not very clear). You probably know that the expectation value of the Hamiltonian in the ground state is usually given as ##\hbar \omega/2##, that's what you would get with your expression. So they decided to subtract that value from your H H in order to achieve, with their hamiltonian, ##\langle 0 | H | 0 \rangle =0##.

nrqed said:
Good work.

The hamiltonian is actually defined up to an additive constant, that is one can *define* the Hamiltonian to be what you got plus any constant energy. This added constant is arbitrary. It is usually taken to be zero, in which case your expression would be the final answer. But what they did here is to choose the constant in such a way that the expectation value ##\langle 0 | H | 0 \rangle =0## , in other words they wanted the ground state energy to be zero (they should have mentioned that in the question, if they did not, they were not very clear). You probably know that the expectation value of the Hamiltonian in the ground state is usually given as ##\hbar \omega/2##, that's what you would get with your expression. So they decided to subtract that value from your H H in order to achieve, with their hamiltonian, ##\langle 0 | H | 0 \rangle =0##.
Cool, thanks for the explanation really helped out.

1. What is a Hamiltonian of a 1D Linear Harmonic Oscillator?

The Hamiltonian of a 1D Linear Harmonic Oscillator is a mathematical operator that represents the total energy of a system consisting of a particle moving back and forth in a linear path under the influence of a force that is directly proportional to its displacement from a fixed point.

2. How is the Hamiltonian of a 1D Linear Harmonic Oscillator calculated?

The Hamiltonian of a 1D Linear Harmonic Oscillator is calculated by adding the kinetic energy operator and the potential energy operator. The kinetic energy operator is equal to the mass of the particle multiplied by its velocity squared, while the potential energy operator is equal to one-half times the spring constant multiplied by the square of the displacement of the particle from the equilibrium position.

3. What is the significance of the Hamiltonian in quantum mechanics?

In quantum mechanics, the Hamiltonian is a fundamental operator that represents the total energy of a system. It is used to describe the time evolution of a quantum state and to calculate the probabilities of different outcomes of measurements. The Hamiltonian of a 1D Linear Harmonic Oscillator is particularly important because many physical systems can be approximated as harmonic oscillators.

4. How does the Hamiltonian of a 1D Linear Harmonic Oscillator change with different parameters?

The Hamiltonian of a 1D Linear Harmonic Oscillator is dependent on three main parameters: the mass of the particle, the spring constant, and the displacement of the particle from the equilibrium position. Changing any of these parameters will alter the Hamiltonian and therefore affect the energy of the system.

5. Can the Hamiltonian of a 1D Linear Harmonic Oscillator be used to solve for the motion of the particle?

Yes, the Hamiltonian of a 1D Linear Harmonic Oscillator can be used to solve for the motion of the particle. By solving the Schrödinger equation, which is based on the Hamiltonian, the wave function of the particle can be determined. This wave function can then be used to calculate the probability of the particle being in a certain position or having a certain energy.

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