# Harmonic oscillator expectation values

• land
In summary, the expectation value of the position, momentum, and total energy would be zero if the harmonic oscillator is in a state that is a superposition of the ground and first excited stationary states.
land
I'm given that there is a harmonic oscillator in a state that is a superposition of the ground and first excited stationary states given by $$\psi = \frac{1}{\sqrt{2}}\abs{\psi_0(x,t) + \psi_1(x,t)}$$, where $$\psi_0 = \psi_0(x)e^{\frac{-iE_0t}{\hbar}}$$ and $$\psi_1 = \psi_1(x)e^{\frac{-iE_1t}{\hbar}}$$. I need to calculate expectation values for position, momentum, and total energy.

Here's what I've done: I'm assuming this is a simple harmonic oscillator, and for the x operator I have $$\frac{i}{\sqrt{2m\omega}}(a_- - a_+)$$. I think that a- operating on Psi0 would be zero, and on Psi1 would be Psi0, and a+ operating on Psi0 would be Psi1, and on Psi1 would be Psi2. So... I have for <x>:

$$<x> = \frac{i}{2\sqrt{2m\omega}}<\psi_0 + \psi_1 | \psi_0 - \psi_1 - \psi_2>$$.. but.. uh.. how do you do this? I feel like I must have done something wrong. A push in the right direction would be much appreciated.

Thanks so much!

There's no way that expectation value could be zero, right? because $$<\psi_n | \psi_{n+1}> = 0$$, I think.. so all that stuff inside the <> would be zero. No?

land said:
I'm given that there is a harmonic oscillator in a state that is a superposition of the ground and first excited stationary states given by $$\psi = \frac{1}{\sqrt{2}}\abs{\psi_0(x,t) + \psi_1(x,t)}$$

Is that supposed to be

$$\Psi = \frac{1}{\sqrt{2}} \left( \abs{\Psi_0(x,t) + \Psi_1(x,t)} \right)?$$

Brackets are important, as are distinctions between upper case and lower case. For example,

$$\psi_0 = \psi_0(x)e^{\frac{-iE_0t}{\hbar}}$$

doesn't make sense without such a distinction. If fact, in this case, it doesn't make sense to write

$$\psi = \psi_{0} + \psi_{1},$$

since the sum of two energy eigenstates is not an energy eigenstate.

and on Psi1 would be Psi2.

No - don't forget the numerical factor.

Now calculate

$$\left< \Psi \left| x \right | \Psi \right>$$

by, as you did, replacing $x$ with the appropriate, but don't to forget to include all the time dependences and other numerical factors.

Also, use the orthogonality condition that you gave in your last post.

I don't mean to sound so critical, but, in this calculation, it's important to get the details right.

I did mean to put brackets around it. I'm not sure why I didn't. I have trouble getting LaTeX to display things correctly sometimes.

Edit: nevermind, I'm an idiot and figured out what you were talking about. See next post for confusion.

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OK. I tried to go back through and put in the terms you were talking about, and I've gotten something.. but I'm still confused. I've gotten down to $$\frac{i}{2\sqrt{2m\omega}}<\psi_0 + \psi_1 | \sqrt{\hbar\omega}\psi_0 - \sqrt{\hbar\omega}\psi_1 - \sqrt{2\hbar\omega}\psi_2>$$. $$<\psi_0 | \psi_0>$$ I know how to do, but $$<\psi_0 | \psi_1>$$ I don't. So you said to use the orthoganility condition I brought up in my last post.. so is this just zero since it's Psi0 and Psi1? If not, here's what I'm getting for that, plugging in the x and t dependences: $$-\sqrt{\hbar\omega}<\psi_0(x)e^{\frac{-iE_0t}{\hbar}} | \psi_1(x)e^{\frac{-iE_1t}{\hbar}}>$$. I have no idea how to do that. So I hope it's zero.

So if it is, then the <Psi0 | Psi0> part of that would just be $$\sqrt{\hbar\omega}$$. That seems way too easy though.

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## 1. What is a harmonic oscillator?

A harmonic oscillator is a system that experiences a restoring force proportional to its displacement from equilibrium. This results in a repetitive motion around a central point.

## 2. How are expectation values calculated for a harmonic oscillator?

The expectation value for a physical quantity in a harmonic oscillator can be calculated by taking the integral of the quantity multiplied by the wave function squared over all possible values.

## 3. What is the significance of harmonic oscillator expectation values?

Harmonic oscillator expectation values provide important information about the average behavior of a system. They can be used to predict the behavior of the system and compare it to experimental results.

## 4. How do expectation values change with different energy levels in a harmonic oscillator?

In a harmonic oscillator, the expectation values for position and momentum increase as the energy level increases. However, the expectation value for energy itself remains constant.

## 5. Can harmonic oscillator expectation values be measured experimentally?

Yes, harmonic oscillator expectation values can be measured experimentally through techniques such as spectroscopy, where the energy levels and corresponding expectation values can be observed and analyzed.

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