Quantum harmonic oscillator: ladder operators

If yes, you can use it, it makes things easier. In summary, the conversation is discussing the evaluation of various values (such as position, momentum, and uncertainty) for a particle in a harmonic oscillator potential. The position and momentum operators can be expressed in terms of ladder operators, which can be used to find the values for different states of the particle. The conversation also mentions using a general formula for n and the use of bra-ket notation or hermitian conjugates to simplify the calculations.
  • #1
tjackson3
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Homework Statement



For a particle of mass m moving in the potential [tex]V(x) = \frac{1}{2}m\omega^2x^2[/tex] (i.e. a harmonic oscillator), it is often convenient to express the position and momentum operators in terms of the ladder operators [tex]a_{\pm}[/tex]:

[tex]x = \sqrt{\frac{\hbar}{2m\omega}}(a_+ + a_-)[/tex]

[tex]p = i\sqrt{\frac{\hbar m \omega}{2}}(a_+ - a_-)[/tex]

Evaluate [tex]<x>,<x^2>,\sigma_x,<p>,<p^2>,\sigma_p[/tex] for the stationary state [tex]\psi_n[/tex], where n = 0 is the ground state, and check that the uncertainty principle is obeyed.

Note that I'm not asking you guys to help with all six of those; I think just a nudge in the right direction on one of them is all I need. Also note that, according to my professor, it is entirely possible to fully complete this problem without evaluating a single integral.

Homework Equations



[tex]\psi_0 = A_0e^{-\frac{m\omega}{2\hbar}x^2}[/tex]

[tex]\psi_n = A_n(a_+)^ne^{-\frac{m\omega}{2\hbar}x^2}[/tex]

[tex]A_n = \left(\frac{m\omega}{\pi\hbar}\right)^{1/4} \frac{(-i)^n}{\sqrt{n!(\hbar\omega)^n}}[/tex]

[tex]a_{\pm} = \frac{1}{\sqrt{2m}}\left(\frac{\hbar}{i}\frac{d}{dx} \pm im\omega x\right)[/tex]

[tex]a_-a_+ - a_+a_- = \hbar\omega[/tex]

[tex]a_-\psi_0 = 0[/tex]

[tex]a_+\psi_n = i\sqrt{(n+1)\hbar\omega}\psi_{n+1}[/tex]

[tex]a_-\psi_n = -i\sqrt{n\hbar\omega}\psi_{n-1}[/tex]

And using this,

[tex]\psi_n = A_n(a_+)^ne^{-\frac{m\omega}{2\hbar}x^2}[/tex]

The Attempt at a Solution



All I'm asking for help on is finding <x>. I think if I can do that, the rest will come fairly easily (if not quickly). At first I tried to do it in the most brute force way possible, using the last equation there in section 2, and literally taking [tex]\psi_n^*x\psi_n[/tex], but that went nowhere, due to the computational difficulties and the difficulty in taking the complex conjugate of that expression in a nice, closed form.

The more promising route seemed to be to find <x> for n = 0 and to use that somehow to generate <x> for some general n. <x> = 0 for n = 0 by parity. Unsure of where to go from there, I tried to use that to find <x> at n = 1. Thus I have (note that [tex]\psi^*_0 = \psi_0[/tex]):

[tex]\[a_+\psi_0 = i\sqrt{\hbar\omega}\psi_1\]

\[\psi_1 = \frac{a_+\psi_0}{i\sqrt{\hbar\omega}}\]\[\psi_1^*x\psi_1 = \frac{a^*_+\psi_0}{-i\sqrt{\hbar\omega}}\sqrt{\frac{\hbar}{2m\omega}}(a_+ + a_-)\frac{a_+\psi_0}{i\sqrt{\hbar\omega}}\][/tex]

I'm stuck here, though, for a couple of reasons. Since operators don't necessarily commute, I'm not sure exactly what liberties I can take in rearranging this into a more convenient form. Additionally, since [tex]a_+[/tex] is an operator, I don't know if it even makes sense to talk about its complex conjugate. Also, short of induction, I don't know how to use this to come up with a formula for a general n.

Thank you so much!
 
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  • #2
Your second attempt is in the right direction. It is sensible to talk about the conjugate of an operator. You only have to be careful with the order
[tex] (a_+\psi)^*= \psi^* a_+^*[/tex]
If you look at the definition of [tex]a_+[/tex] you will find it's conjugate. Then using the commutation relations for the a's you will find a formulat for general n.

By the way, did your professor mention bra-ket notation or hermitian conjugate in the lecture?
 

1. What is a quantum harmonic oscillator?

A quantum harmonic oscillator is a physical system that exhibits simple harmonic motion, similar to a mass on a spring. However, in quantum mechanics, the system is described by a wave function and the energy levels are quantized.

2. What are ladder operators in the context of a quantum harmonic oscillator?

Ladder operators in the context of a quantum harmonic oscillator are mathematical operators that are used to raise or lower the energy level of the system. They are used to describe the transitions between energy levels and are essential in solving the quantum harmonic oscillator problem.

3. How do ladder operators work?

Ladder operators work by acting on the wave function of the quantum harmonic oscillator. The raising operator increases the energy level by one unit, while the lowering operator decreases the energy level by one unit. These operators follow certain mathematical rules and can be used to find the energy states of the quantum harmonic oscillator system.

4. What is the significance of ladder operators in quantum mechanics?

Ladder operators are significant in quantum mechanics because they allow us to understand the behavior of quantum systems, such as the quantum harmonic oscillator. They also help us to calculate the probabilities of different energy states and transitions between them.

5. Are ladder operators only applicable to quantum harmonic oscillators?

No, ladder operators are also applicable to other quantum mechanical systems, such as the hydrogen atom. They are a fundamental tool in solving many quantum mechanical problems and understanding the behavior of quantum systems.

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