# Last year's exam paper

30/01/14 23:36

Last year’s TP2 paper is here. And here are the worked solutions. At this point we have covered the material needed to answer the first two questions.

Older exam papers, from the previous incarnation of this course, are still available on Nigel Cooper’s web page. Scattering theory is unchanged from that course, and there is significant overlap in the treatment of density matrices. So questions on these two topics from the older course are worth working through.

If you need more problems, please bear in mind that every one of the topics in this course is covered in almost all advanced texts in quantum mechanics. In particular, this book contains more than 700 solved problems. No need to go hungry!

I was an undergraduate here, and recall rarely opening any of the recommended books. What was I thinking? Course handouts are a good start but leave a false impression that any subject is bounded and finite. Your friends in the humanities read dozens of books per term and don’t suffer the same delusion.

Quantum mechanics is the same subject whether in Cambridge or Cantabria. If you find one treatment of a topic confusing, seek out one you like better.

Older exam papers, from the previous incarnation of this course, are still available on Nigel Cooper’s web page. Scattering theory is unchanged from that course, and there is significant overlap in the treatment of density matrices. So questions on these two topics from the older course are worth working through.

If you need more problems, please bear in mind that every one of the topics in this course is covered in almost all advanced texts in quantum mechanics. In particular, this book contains more than 700 solved problems. No need to go hungry!

I was an undergraduate here, and recall rarely opening any of the recommended books. What was I thinking? Course handouts are a good start but leave a false impression that any subject is bounded and finite. Your friends in the humanities read dozens of books per term and don’t suffer the same delusion.

Quantum mechanics is the same subject whether in Cambridge or Cantabria. If you find one treatment of a topic confusing, seek out one you like better.

Comments

# More on the Berry phase

30/01/14 20:27

I know that students find their first encounter with the Berry phase quite a challenge. On the other hand, I’m guessing that the adiabatic approximation that we discuss first seems a lot more straightforward. My aim in this note is to show that that this apparent difference in difficulty is illusory. I suppose this may be good or bad news....

Recall that the idea of the adiabatic approximation was to express the state of our system in terms of the

$$|\Psi(t)\rangle = c_{+}(t)|+_{t}\rangle+c_{-}(t)|-_{t}\rangle$$

One then finds that in the adiabatic approximation (when the Hamiltonian changes sufficiently slowly -- see the notes for the precise formulation) the evolution of the coefficients under the time dependent Schrödinger equation is very simple

$$c_{\pm}(t) = \exp\left(-\frac{i}{\hbar}\int_{0}^{t}E_{\pm}(t')\, dt'\right)c_{\pm}(0).$$

This is very appealing, as it seems like a modest generalisation of the evolution of stationary states. Unfortunately, it is meaningless! The problem is that, since \(|\pm_{t}\rangle\) are defined as eigenstates, their phase is not fixed. We could just as well take some states differing by arbitrary phases at each instant in time. Now bear in mind that we are searching for a solution of the time dependent Schrödinger equation, completely determined (including its phase) by the initial state. So this arbitrariness will not do!

The solution is to fix once and for all the phase of the eigenstates for each value of the instantaneous Hamiltonian. That’s why we switch to labelling them by \(\mathbf{H}\) rather than time. This seems rather modest, but it has dramatic consequences.

In our original derivation of the adiabatic approximation, we assumed (tacitly) that as the Hamiltonian changed by a differential amount, we only had to account for the change of the eigenstate in an orthogonal direction. If we were working with real vectors, this would be fairly clear, as the change of a normalised vector can only be perpendicular to the vector. For complex normalised vectors, we have the phase to worry about, but we ignored this subtlety on the first pass.

It turns out that it is not possible to choose the states so that the change in the states is only perpendicular to the states. The best way to convince yourself of this is to actually find the eigenstates explicitly in terms of the direction of \(\mathbf{H}\) that defines the Hamiltonian. There are many ways to do this, corresponding to different assignments of phase, but one way is done in Eq. (2.37-2.39). Then you find

$$\langle \mathbf{H},+|(\delta|\mathbf{H},+\rangle)=i \mathbf{A}_+\cdot\delta \mathbf{H},$$

which will be nonzero. Any other assignment of the phase to the eigenstates can only lead to a gauge transformation of \(\mathbf{A}_+\). Such a transformation can never eliminate \(\mathbf{A}_+\) altogether. One way to see that is to find the curl of \(\mathbf{A}_+\). It is nonzero, and can’t be changed by a gauge transformation.

In particular, the total phase change (in addition to the dynamical phase involving the energy discussed above) is the line integral of \(\mathbf{A}_+\) around a closed loop in parameter space. It is related to the integral of the curl over a bounding surface, by Stokes’ theorem, and gives a gauge invariant phase change:

As I say in the notes, I do recommend looking at Berry’s paper

Recall that the idea of the adiabatic approximation was to express the state of our system in terms of the

*instantaneous eigenbasis*of the Hamiltonian. For the spin-1/2 system discussed in the lectures, this is$$|\Psi(t)\rangle = c_{+}(t)|+_{t}\rangle+c_{-}(t)|-_{t}\rangle$$

One then finds that in the adiabatic approximation (when the Hamiltonian changes sufficiently slowly -- see the notes for the precise formulation) the evolution of the coefficients under the time dependent Schrödinger equation is very simple

$$c_{\pm}(t) = \exp\left(-\frac{i}{\hbar}\int_{0}^{t}E_{\pm}(t')\, dt'\right)c_{\pm}(0).$$

This is very appealing, as it seems like a modest generalisation of the evolution of stationary states. Unfortunately, it is meaningless! The problem is that, since \(|\pm_{t}\rangle\) are defined as eigenstates, their phase is not fixed. We could just as well take some states differing by arbitrary phases at each instant in time. Now bear in mind that we are searching for a solution of the time dependent Schrödinger equation, completely determined (including its phase) by the initial state. So this arbitrariness will not do!

The solution is to fix once and for all the phase of the eigenstates for each value of the instantaneous Hamiltonian. That’s why we switch to labelling them by \(\mathbf{H}\) rather than time. This seems rather modest, but it has dramatic consequences.

In our original derivation of the adiabatic approximation, we assumed (tacitly) that as the Hamiltonian changed by a differential amount, we only had to account for the change of the eigenstate in an orthogonal direction. If we were working with real vectors, this would be fairly clear, as the change of a normalised vector can only be perpendicular to the vector. For complex normalised vectors, we have the phase to worry about, but we ignored this subtlety on the first pass.

It turns out that it is not possible to choose the states so that the change in the states is only perpendicular to the states. The best way to convince yourself of this is to actually find the eigenstates explicitly in terms of the direction of \(\mathbf{H}\) that defines the Hamiltonian. There are many ways to do this, corresponding to different assignments of phase, but one way is done in Eq. (2.37-2.39). Then you find

$$\langle \mathbf{H},+|(\delta|\mathbf{H},+\rangle)=i \mathbf{A}_+\cdot\delta \mathbf{H},$$

which will be nonzero. Any other assignment of the phase to the eigenstates can only lead to a gauge transformation of \(\mathbf{A}_+\). Such a transformation can never eliminate \(\mathbf{A}_+\) altogether. One way to see that is to find the curl of \(\mathbf{A}_+\). It is nonzero, and can’t be changed by a gauge transformation.

In particular, the total phase change (in addition to the dynamical phase involving the energy discussed above) is the line integral of \(\mathbf{A}_+\) around a closed loop in parameter space. It is related to the integral of the curl over a bounding surface, by Stokes’ theorem, and gives a gauge invariant phase change:

*Berry’s phase.*As I say in the notes, I do recommend looking at Berry’s paper

# The Dyson series

23/01/14 16:05

Several people asked to see a route from the Schrödinger equation

\[i\hbar \frac{\partial |\psi(t)\rangle}{\partial t}=H(t)|\psi(t)\rangle \]

to the Dyson series for the evolution operator

\[ U(t,t') = 1 -\frac{i}{\hbar}\int_{t'}^{t} H(t_1)\, dt_{1} -\frac{1}{\hbar^{2}} \int_{t'}^{t}dt_{1}\int_{t’}^{t_{1}}dt_{2}\, H(t_{1})H(t_{2})+\cdots.\]

Since I see that this is confined to a footnote of the Advanced Quantum course, here is the derivation. Integrate both sides of the Schrödinger equation with respect to time to give

\[|\psi(t)\rangle=|\psi(t’)\rangle-\frac{i}{\hbar}\int_{t’}^{t} H(t_1)|\psi(t_1)\rangle dt_1\]

and then iterate. You will see that at each stage of the iteration, the new time integral you introduce goes from the initial time \(t’\) to the previous dummy variable. In other words, the occurrences of the \(H(t)\) are

Finally, what is the relation to the exponential series?

\[\mathcal{T} \exp\left(-\frac{i}{\hbar}\int_{t'}^{t} H(t_{i})\, dt_{i}\right)\]

The exponential series of course has a factor \(1/n!\) in the \(n^\text{th}\) order. Where does this come from? This is because restricting the dummy variables to an ordering \(t_1>t_2>\ldots t_n\) restricts the n-dimensional volume of the integral by a factor \(1/n!\) (think of the number of ways of permuting the labels).

\[i\hbar \frac{\partial |\psi(t)\rangle}{\partial t}=H(t)|\psi(t)\rangle \]

to the Dyson series for the evolution operator

\[ U(t,t') = 1 -\frac{i}{\hbar}\int_{t'}^{t} H(t_1)\, dt_{1} -\frac{1}{\hbar^{2}} \int_{t'}^{t}dt_{1}\int_{t’}^{t_{1}}dt_{2}\, H(t_{1})H(t_{2})+\cdots.\]

Since I see that this is confined to a footnote of the Advanced Quantum course, here is the derivation. Integrate both sides of the Schrödinger equation with respect to time to give

\[|\psi(t)\rangle=|\psi(t’)\rangle-\frac{i}{\hbar}\int_{t’}^{t} H(t_1)|\psi(t_1)\rangle dt_1\]

and then iterate. You will see that at each stage of the iteration, the new time integral you introduce goes from the initial time \(t’\) to the previous dummy variable. In other words, the occurrences of the \(H(t)\) are

*time ordered*from later to earlier times as you read from left to right. The symbol \(\mathcal{T}\) just tells us to do that.Finally, what is the relation to the exponential series?

\[\mathcal{T} \exp\left(-\frac{i}{\hbar}\int_{t'}^{t} H(t_{i})\, dt_{i}\right)\]

The exponential series of course has a factor \(1/n!\) in the \(n^\text{th}\) order. Where does this come from? This is because restricting the dummy variables to an ordering \(t_1>t_2>\ldots t_n\) restricts the n-dimensional volume of the integral by a factor \(1/n!\) (think of the number of ways of permuting the labels).

# TP2: Topics in Quantum Theory

23/01/14 16:03

The lecture handout for TP2 went to press on the 14th January. Any corrections, improvements, or additions (non examinable, of course) will be posted here.

Please let me know of any errors you spot.