For my homework i am considering a harmonic oscillator which´s wavefunction at $t=0$ is the superposition of the eigenstates $\psi_n$. $$ \psi(x,t=0) = \sum\nolimits_{n} c_n \cdot \psi_n(x) $$ Now i am asked for the probability of the oscillator occupying eigenenergies $E_n > 2\hbar\omega$ at some point in time which is, taking the energy formula for the harmonic oscillator $E_n = \hbar\omega(n+\frac{1}{2})$ into consideration, equivalent to asking for the probability of the two lowest states. So how do I calculate them? Usually I would do it via the orthogonality relation of the eigenstates but in this case i am missing a kind of boundary condition which i could plug in for my superposition.
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$\begingroup$ Wouldn't you just pay attention to the coefficients of the first two values in the sum? $\endgroup$– roshokaJun 2, 2020 at 21:55
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$\begingroup$ Yes. However i don´t know how to calculate either. $\endgroup$– minitsJun 2, 2020 at 22:24
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2 Answers
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The answer provided by roshoka is correct. I only want to briefly discuss why the time dependence can be "ignored" in this case.
As the question asks for the probability of obtaining a particular energy eigenvalue at a later time, then you must first understand what the wave function looks like at a later time t. For a system in which the potential is time independent (like the harmonic oscillator), then you can solve the time dependence of the Schrödinger equation once and for all. If you do that, you find that each energy eigenstate evolves independently according to $e^{-iE_nt/\hbar}$, so that you can write your wave function at time $t$ as: $$ \Psi(x,t)=\sum_nc_n\psi_n(x)e^{-iE_nt/\hbar}. $$ For a harmonic oscillator, you are correct that $E_n=\hbar\omega\left(n+\frac{1}{2}\right)$, so that energies $E_n>2\hbar\omega$ correspond to $n>1$. This means that the total probability of measuring an energy $E_n>2\hbar\omega$ is given by the probability of measuring each eigenvalue for $n>1$. The probability of measuring a particular eigenvalue $E_n$ is: $$ P(E_n)=\left|\int_{-\infty}^{\infty}\psi_n^{\ast}(x)\Psi(x,t)dx\right|^2=\left|\int_{-\infty}^{\infty}\psi_n^{\ast}(x)\left(\sum_mc_m\psi_m(x)e^{-iE_mt/\hbar}\right)dx\right|^2= \left|\sum_mc_me^{-iE_mt/\hbar}\int_{-\infty}^{\infty}\psi_n^{\ast}(x)\psi_m(x)dx\right|^2= \left|\sum_mc_me^{-iE_mt/\hbar}\delta_{nm}\right|^2 =|c_ne^{-iE_nt/\hbar}|^2=|c_n|^2. $$ To solve the integral, I have used the orthonormality of the eigenstates. Therefore, the total probability of measuring the system in an energy eigenstate $E_n>2\hbar\omega$ is: $$ P(E_n>2\hbar\omega)=\sum_{n=2}^{\infty}|c_n|^2, $$ note that the sum starts at $n=2$. This can be more simply calculated as: $$ P(E_n>2\hbar\omega)=1-\left(|c_0|^2+|c_1|^2\right), $$ as roshoka suggests. All of this analysis assumes that the original wave function is normalized.
It may seem redundant having to first construct the wave function at a later time $\Psi(x,t)$ given that the final answer did not depend on time. However, this arises only because you are interested in an energy measurement and the dynamics of the system is governed by the energy eigenstates (this is because the operator that appears in the Schrödinger equation is the Hamiltonian). For the measurement of any other property, the time dependence would not be trivial like in this case, but the approach above would still work.
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$\begingroup$ Just commenting the dirac version with propogator (partially for my own sake)$|\Psi(t)\rangle = U(t)|\Psi(0)\rangle = \sum_n |n\rangle\langle n|\Psi(0)\rangle e^{\frac{-i E_n t}{\hbar}}$ then $|\langle j|\Psi(t)\rangle|^2 = |\sum_n \langle j|n\rangle\langle n|\Psi(0)\rangle e^{\frac{-i E_n t}{\hbar}}|^2 = |\sum_n \delta_{j,n}\langle n|\Psi(0)\rangle e^{\frac{-i E_n t}{\hbar}}|^2 = |\langle j |\Psi(0)\rangle e^{\frac{-i E_j t}{\hbar}}|^2 = |\langle j|\Psi(0)\rangle|^2 = |c_j|^2$ $\endgroup$– roshokaJun 4, 2020 at 2:19
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A fundamental postulate of QM is (From Shankar chpt. 4)
If the particle is in a state $|\psi\rangle$, measurement of the variable (corresponding to) $\Omega$ will yield one of the eigenvalues $\omega$ with proabability $P(\omega_i) \propto|\langle\omega_i|\psi\rangle|^2$. And $P(\omega_i) = |\langle\omega_i|\psi\rangle|^2$ if $|\psi\rangle$ is normalized.
Your eigenvalues are $E_n$, and your eigenvectors are $|E_n\rangle$. So what you need to do is find the complement of the sum of the probabilities for $n=0,1$ using the above.
Do you have actual values or does it need to be done symbolically?
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$\begingroup$ No, i don´t have any real values to use. Actually, in my course we are just one lecture away from introducing dirac notation. Therefore, is my interpretation right? $P(\omega_i) = \int \overline{\psi(x)}\cdot E_n \cdot \psi(x) \ dx $ $\endgroup$– minitsJun 3, 2020 at 0:37
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$\begingroup$ You could think of it like $P(E_n) = |\langle E_n|\psi\rangle|^2 = \int |\psi_n^*(x) \Psi(x)|^2dx $ then just use orthogonality. Where $\psi_n$ is the eigenfunction corresponding to $E_n$. In the end you should get squares of coefficients, which are probabilities. Take the complement of the sum for $n=0,1$ $\endgroup$– roshokaJun 3, 2020 at 1:55
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$\begingroup$ --The above comment should have $|\langle E_n | \Psi \rangle |^2$. Where $\Psi$ is the function from your question.-- You could kind of just hand wave it and think about it like multiplying $\psi_0$ into the sum you have in your question and use the orthogonality (and don't forget to square it). Do the same for $\psi_1$. The answers to both of these will be their individual probabilities. Then use those to find out what the probability is for $n>2$ $\endgroup$– roshokaJun 3, 2020 at 2:07
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$\begingroup$ Another edit: In my first comment, the absolute value squared should be on the outside of the integral. $\endgroup$– roshokaJun 4, 2020 at 2:09