The instructor gives a brief lecture about time dependence of energy eigenstates (e.g. McIntyre, 3.1). Notes for the students are attached.
1. << Superposition States for a Particle on a Ring | Quantum Ring Sequence | Time Dependence for a Quantum Particle on a Ring >>
Begin with Schrödinger's Equation: \begin{equation} i\hbar\frac{d}{dt}\left|{\Psi(t)}\right\rangle =\hat{H}\left|{\Psi(t)}\right\rangle \end{equation} If we know how the Hamiltonian acts on a state, we can use Schrodinger's equation to find out how it evolves in time.
Let's do this for the case of the particle on a ring (c.f. McIntyre 3.1 for a spin 1/2 system). Write out the arbitrary state as a superposition of energy eigenstates, putting the time dependence we are trying to find in the expansion coefficient. \begin{equation} i\hbar\frac{d}{dt}\sum_{m=-\infty}^{\infty} c_m(t) \left|{m}\right\rangle =\hat{H}\sum_{m=-\infty}^{\infty} c_m(t) \left|{m}\right\rangle \end{equation} We are writing our state in this basis because we know what the Hamiltonian does to these kets \(\left|{m}\right\rangle \) via the energy eigenvalue equation. \begin{equation} \hat{H}\left|{m}\right\rangle =E_m\left|{m}\right\rangle \end{equation} Using this result, we can calculate: \begin{align} i\hbar\frac{d}{dt}\sum_{m=-\infty}^{\infty} c_m(t) \left|{m}\right\rangle &=\hat{H}\sum_{m=-\infty}^{\infty} c_m(t) \left|{m}\right\rangle \\ &=\sum_{m=-\infty}^{\infty} c_m(t) \hat{H}\left|{m}\right\rangle \\ &=\sum_{m=-\infty}^{\infty} E_m c_m(t) \left|{m}\right\rangle \\ &=\sum_{m=-\infty}^{\infty}\left(\frac{\hbar^2}{2I}\, m^2\right)\, c_m(t) \left|{m}\right\rangle \label{energy} \end{align} The energies \(E_m\) are the energies for the physical system we care about, i.e. a particle on a ring. If we had a different system, a different Hamiltonian, the structure of this equation would be similar but the values of energy in Eqn. (\ref{energy}) would be different.
Because we are solving for the time-dependent coefficients, let's get rid of the sum using an inner product with an arbitratry state \(\left\langle {k}\right|\) by using the orthonormality condition. \begin{align} \left\langle {k}\right|i\hbar\frac{d}{dt}\sum_{m=-\infty}^{\infty} c_m(t) \left|{m}\right\rangle &=\left\langle {k}\right|\sum_{m=-\infty}^{\infty} E_m c_m(t) \left|{m}\right\rangle \\ i\hbar\frac{d}{dt}\sum_{m=-\infty}^{\infty} c_m(t) \left\langle {k}\middle|{m}\right\rangle &=\sum_{m=-\infty}^{\infty} E_m c_m(t) \left\langle {k}\middle|{m}\right\rangle \\ i\hbar\frac{d}{dt}\sum_{m=-\infty}^{\infty} c_m(t) \delta_{km} &=\sum_{m=-\infty}^{\infty} E_m c_m(t) \delta_{km}\\ i\hbar\frac{d}{dt} c_k(t) &= E_k c_k(t)\\ \end{align} We now have a first-order ODE for the unknown time-dependent coefficients. \begin{equation} \frac{d}{dt}c_k(t)=-\frac{iE_k}{\hbar}c_k(t) \end{equation} Because this ODE is linear with constant coefficients, the solution is an exponential. \begin{equation} c_k(t)=Ae^{-\frac{iE_k}{\hbar}t} \end{equation} The constant factor \(A\) is what we have been calling the time-independent version of \(c_k\) up to this point; it is the probability amplitude of the eigenstate it accompanies at \(t=0\). You can plug in \(t=0\) to help verify this.
The solution is: \begin{align} \left|{\Psi(t)}\right\rangle &=\sum_{m=-\infty}^{\infty} e^{-\frac{iE_m}{\hbar}t}c_m \left|{m}\right\rangle \\ &=\sum_{m=-\infty}^{\infty} e^{-\frac{i\hbar}{2I}\, m^2t}c_m \left|{m}\right\rangle \end{align} Here are some important features of the solution:
- We can only put time evolution phases on states that are written in the energy basis;
- The time dependent phase is pure imaginary; it does not effect the normalization of the state;
- Each eigenstate gets its own time-dependent phase that includes the energy of that eigenstate.
It would be valuable for you to go through this derivation and see which parts depend on the particular system we are using (the quantum ring) and which parts are generic to any solution of Schrödinger's Equation.
Ask students to write on a small whiteboard something they know/remember about Schrödinger's Equation. Use their responses to write the general form: \begin{equation} i\hbar\frac{d}{dt}\left|{\Psi(t)}\right\rangle =\hat{H}\left|{\Psi(t)}\right\rangle \end{equation} If we know how the Hamiltonian acts on a state, we can use Schrodinger's equation to find out how it evolves in time.
Let's do this for the case of the particle on a ring (c.f. McIntyre 3.1 for a spin 1/2 system). Write out the arbitrary state as a superposition of energy eigenstates, putting the time dependence we are trying to find in the expansion coefficient. \begin{equation} i\hbar\frac{d}{dt}\sum_{m=-\infty}^{\infty} c_m(t) \left|{m}\right\rangle =\hat{H}\sum_{m=-\infty}^{\infty} c_m(t) \left|{m}\right\rangle \end{equation} Remind the students that we are writing our state in this basis because we know what the Hamiltonian does to these kets \(\left|{m}\right\rangle \). This could be a good place to ask students to write the energy eigenvalue equation. \begin{equation} \hat{H}\left|{m}\right\rangle =E_m\left|{m}\right\rangle \end{equation} Use their responses to get calculate: \begin{align} i\hbar\frac{d}{dt}\sum_{m=-\infty}^{\infty} c_m(t) \left|{m}\right\rangle &=\sum_{m=-\infty}^{\infty}\hat{H} c_m(t) \left|{m}\right\rangle \\ &=\sum_{m=-\infty}^{\infty} E_m c_m(t) \left|{m}\right\rangle \\ &=\sum_{m=-\infty}^{\infty}\left(\frac{\hbar^2}{2I}\, m^2\right)\, c_m(t) \left|{m}\right\rangle \end{align} Empasize that the energies \(E_m\) are the energies for a particle on a ring. If we had a different system, a different Hamiltonian, the structure of this equation would be similar but the values of energy would be different.
Because we are solving for the time-dependent coefficients, let's try to get rid of the sum using an inner product with an arbitratry state \(\left\langle {k}\right|\) so we can use the orthonormality condition. \begin{align} \left\langle {k}\right|i\hbar\frac{d}{dt}\sum_{m=-\infty}^{\infty} c_m(t) \left|{m}\right\rangle &=\left\langle {k}\right|\sum_{m=-\infty}^{\infty} E_m c_m(t) \left|{m}\right\rangle \\ i\hbar\frac{d}{dt}\sum_{m=-\infty}^{\infty} c_m(t) \left\langle {k}\middle|{m}\right\rangle &=\sum_{m=-\infty}^{\infty} E_m c_m(t) \left\langle {k}\middle|{m}\right\rangle \\ i\hbar\frac{d}{dt}\sum_{m=-\infty}^{\infty} c_m(t) \delta_{km} &=\sum_{m=-\infty}^{\infty} E_m c_m(t) \delta_{km}\\ i\hbar\frac{d}{dt} c_k(t) &= E_k c_k(t)\\ \end{align} We now have a first-order, linear ODE with constant coefficients. \begin{equation} \frac{d}{dt}c_k(t)=-\frac{iE_k}{\hbar}c_k(t) \end{equation} See if the students recall what the solution to this equation is. There should be someone who recognizes the solution to be an exponential. \begin{equation} c_k(t)=Ae^{-\frac{iE_k}{\hbar}t} \end{equation} Explain that the constant factor \(A\) is what they have been calling the time independent version of \(c_k\) up to this point, it is the probability amplitude of the eigenstate it accompanies at \(t=0\). You can have them plug in \(t=0\) to help verify this. \begin{equation} \left|{\Psi(t)}\right\rangle =\sum_m e^{-\frac{iE_m}{\hbar}t}c_m \left|{m}\right\rangle \end{equation} Point out important features of the solution:
assignment Homework
assignment Homework
group Small Group Activity
30 min.
central forces quantum mechanics eigenstates eigenvalues quantum measurements angular momentum hermitian operators probability superposition
Students calculate probabilities for a particle on a ring whose wavefunction is not easily separated into eigenstates by inspection. To find the energy, angular momentum, and position probabilities, students perform integrations with the wavefunction or decompose the wavefunction into a superposition of eigenfunctions.keyboard Computational Activity
120 min.
quantum mechanics operator matrix element particle in a box eigenfunction
Students find matrix elements of the position operator \(\hat x\) in a sinusoidal basis. This allows them to express this operator as a matrix, which they can then numerically diagonalize and visualize the eigenfunctions.assignment Homework
As discussed in class, we can consider a black body as a large box with a small hole in it. If we treat the large box a metal cube with side length \(L\) and metal walls, the frequency of each normal mode will be given by: \begin{align} \omega_{n_xn_yn_z} &= \frac{\pi c}{L}\sqrt{n_x^2 + n_y^2 + n_z^2} \end{align} where each of \(n_x\), \(n_y\), and \(n_z\) will have positive integer values. This simply comes from the fact that a half wavelength must fit in the box. There is an additional quantum number for polarization, which has two possible values, but does not affect the frequency. Note that in this problem I'm using different boundary conditions from what I use in class. It is worth learning to work with either set of quantum numbers. Each normal mode is a harmonic oscillator, with energy eigenstates \(E_n = n\hbar\omega\) where we will not include the zero-point energy \(\frac12\hbar\omega\), since that energy cannot be extracted from the box. (See the Casimir effect for an example where the zero point energy of photon modes does have an effect.)
Show that the free energy is given by \begin{align} F &= 8\pi \frac{V(kT)^4}{h^3c^3} \int_0^\infty \ln\left(1-e^{-\xi}\right)\xi^2d\xi \\ &= -\frac{8\pi^5}{45} \frac{V(kT)^4}{h^3c^3} \\ &= -\frac{\pi^2}{45} \frac{V(kT)^4}{\hbar^3c^3} \end{align} provided the box is big enough that \(\frac{\hbar c}{LkT}\ll 1\). Note that you may end up with a slightly different dimensionless integral that numerically evaluates to the same result, which would be fine. I also do not expect you to solve this definite integral analytically, a numerical confirmation is fine. However, you must manipulate your integral until it is dimensionless and has all the dimensionful quantities removed from it!
Show that the entropy of this box full of photons at temperature \(T\) is \begin{align} S &= \frac{32\pi^5}{45} k V \left(\frac{kT}{hc}\right)^3 \\ &= \frac{4\pi^2}{45} k V \left(\frac{kT}{\hbar c}\right)^3 \end{align}
Show that the internal energy of this box full of photons at temperature \(T\) is \begin{align} \frac{U}{V} &= \frac{8\pi^5}{15}\frac{(kT)^4}{h^3c^3} \\ &= \frac{\pi^2}{15}\frac{(kT)^4}{\hbar^3c^3} \end{align}
assignment Homework
Any sufficiently smooth spatial wave function inside a 2-D box can be expanded in a double sum of the product wave functions, i.e. \begin{equation} \psi(x,y)=\sum_{n=1}^{\infty}\sum_{m=1}^{\infty}\, c_{nm}\; \hbox{eigenfunction}_n(x)\;\hbox{eigenfunction}_m(y) \end{equation} Using your expressions from part (a) above, write out all the terms in this sum out to \(n=3\), \(m=3\). Arrange the terms, conventionally, in terms of increasing energy.
You may find it easier to work in bra/ket notation: \begin{align*} \left|{\psi}\right\rangle &=\sum_{n=1}^{\infty}\sum_{m=1}^{\infty}\, c_{nm}\left|{n}\right\rangle \left|{m}\right\rangle \\ &=\sum_{n=1}^{\infty}\sum_{m=1}^{\infty}\, c_{nm}\left|{nm}\right\rangle \end{align*}
assignment Homework
Find the chemical potential of an ideal monatomic gas in two dimensions, with \(N\) atoms confined to a square of area \(A=L^2\). The spin is zero.
Find an expression for the energy \(U\) of the gas.
Find an expression for the entropy \(\sigma\). The temperature is \(kT\).
assignment Homework
In each of the following sums, shift the index \(n\rightarrow n+2\). Don't forget to shift the limits of the sum as well. Then write out all of the terms in the sum (if the sum has a finite number of terms) or the first five terms in the sum (if the sum has an infinite number of terms) and convince yourself that the two different expressions for each sum are the same:
group Small Group Activity
30 min.
assignment Homework