How do you find the probabilities for an anharmonic quantum oscillator state?

[damarkk]

Homework Statement

Find the probability for all possible unperturbed states $| n\rangle$ for an harmonic oscillator with perturbation $V(x)=\alpha x^3$

Relevant Equations

$x$, $p$.

I have one tremendous doubt about it.


On $t=0$ the state of the oscillator is $| \Psi (t) \rangle = | 1 \rangle$. The perturbation is $V(x)=\alpha x^3 = \alpha (\frac{\hbar}{2m\omega})^{3/2} (a+a^{\dagger})^3 = \gamma (a^3+3Na+3Na^{\dagger} + 3a + (a^{\dagger})^3)$.

The only possible other states are $|0 \rangle$, $| 2\rangle$, $|4\rangle$. The state corrected on the first order is:

$|\Psi \rangle = |1 \rangle + c_0|0 \rangle + c_2|2 \rangle + c_4|4\rangle$

where $c_k = \langle k| V(x) | 1 \rangle$.


What are the probability to find the oscillator in a generic state $|n \rangle$?


My answer is $P_k = |c_k|^2$, but for k=1 we have $P_1 = 1$. This is not possible I think. On the other hand, I suppose that $P_1 = 1-P_0 - P_2- P_4 = 1-|c_0|^2-|c_2|^2-|c_4|^2$, but how can I show this? The coefficient of $| 1\rangle$ is 1. This is my question.

 

[TSny]

Homework Statement: Find the probability for all possible unperturbed states $| n\rangle$ for an harmonic oscillator with perturbation $V(x)=\alpha x^3$
Relevant Equations: $x$, $p$.

On $t=0$ the state of the oscillator is $| \Psi (t) \rangle = | 1 \rangle$.

How do you know the state is $|1\rangle$ at $t = 0$? This was not given in the homework statement as given. Always provide the full homework statement.

I don't understand "$x$, $p$" for your Relevant Equations.

The perturbation is $V(x)=\alpha x^3 = \alpha (\frac{\hbar}{2m\omega})^{3/2} (a+a^{\dagger})^3 = \gamma (a^3+3Na+3Na^{\dagger} + 3a + (a^{\dagger})^3)$.

The only possible other states are $|0 \rangle$, $| 2\rangle$, $|4\rangle$. The state corrected on the first order is:

$|\Psi \rangle = |1 \rangle + c_0|0 \rangle + c_2|2 \rangle + c_4|4\rangle$

where $c_k = \langle k| V(x) | 1 \rangle$.

Your expression for $c_k$ doesn't have the correct dimensions. Did you leave out something?

What are the probability to find the oscillator in a generic state $|n \rangle$?


My answer is $P_k = |c_k|^2$, but for k=1 we have $P_1 = 1$. This is not possible I think. On the other hand, I suppose that $P_1 = 1-P_0 - P_2- P_4 = 1-|c_0|^2-|c_2|^2-|c_4|^2$, but how can I show this? The coefficient of $| 1\rangle$ is 1. This is my question.

From $|\Psi \rangle = |1 \rangle + c_0|0 \rangle + c_2|2 \rangle + c_4|4\rangle$, you can see that $|\Psi \rangle$ is not normalized.

That's ok. The probability $P_k$ is then $|c_k|^2/\langle \Psi |\Psi \rangle$.