Monthly Archive: January 2015

In the last two posts, we’ve discussed the path integral and how quantum evolution can be thought of as having contributions from every possible path in space-time such that the sum of their contributions exactly defines the quantum evolution operator $$U$$. In addition, we found that potentials in one dimension of the form $$V = a + b x + c x^2 + d \dot x + e x \dot x$$ kindly cooperate with the evaluation of the path integral. While potentials of these types do lend themselves to problems of both practical and theoretical importance, they exclude one very important class of problems – namely time-dependent potentials. Much of our modern economy is built upon time-dependent electric and magnetic fields, including the imaging sciences of photography and motion pictures, medical and magnetic resonance imaging, microwave ovens, modern electronics, and many more. In this post, I’ll be discussing the general structure for calculating how a quantum state evolves under a time-varying force. The main ingredients in the procedure are the introduction of a new picture, similar to the Schrodinger and Heisenberg pictures, and the perturbative expansion in this picture of the quantum evolution operator.

We start by assuming that the Hamiltonian can be written as
\[ H = H_0 + V(t) \; ,\]

where $$H_0$$ represents the Hamiltonian for some model problem that we can solve exactly. Usually $$H_0$$ represents the free-particle case.

Obviously, the aim is to solve the old and familiar state evolution equation
\[ i \hbar \frac{d}{dt} \left| \psi(t) \right> = H \left| \psi(t) \right> \]
to get the evolution operator that connects the state at the initial time $$t_0$$ with the state at time $$t$$
\[ \left| \psi(t) \right> = U(t,t_0) \left| \psi(t_0) \right> \; .\]
Since we haven’t nailed down any of the attributes of our model Hamiltonian other than it be exactly solvable, I can assume $$H_0 \neq H_0(t)$$. With this assumption, the evolution operator corresponding to $$H_0$$ then
becomes
\[ U_0(t,t_0) = e^{-i H_0(t – t_0)/\hbar} \; , \]
and its inverse is given by the Hermitian conjugate
\[U_0^{-1}(t,t_0) = e^{i H_0(t – t_0)/\hbar} \; .\]

The next step is to introduce a new state $$\left| \lambda (t) \right>$$ defined through the relation
\[ \left| \psi(t) \right> = U_0(t,t_0) \left| \lambda(t) \right> \; .\]
An obvious consequence of the above relation is the boundary condition
\[ \left| \psi(t_0) \right> = \left| \lambda(t_0) \right> \]
when $$t = t_0$$. This relation will come to be useful later.

By introducing this state, we’ve effectively introduced a new picture in which the state kets are defined with respect to a frame that ‘rotates’ in step with the evolution caused by $$H_0$$. This picture is called the Interaction or Dirac picture.

The evolution of this state obeys
\[ \frac{d}{dt} \left| \lambda(t) \right> = \frac{i}{\hbar} H_0 e^{i H_0(t – t_0)/\hbar} \left| \psi(t) \right> + e^{i H_0(t – t_0)/\hbar} \frac{d}{dt} \left| \psi(t) \right> \; , \]
which, when substituting the right-hand side of the time evolution of $$\left| \psi(t) \right>$$, simplifies to
\[ \frac{d}{dt} \left| \lambda(t) \right> = \frac{1}{i\hbar} e^{i H_0(t – t_0)/\hbar} \left[H – H_0\right] \left| \psi(t) \right> \; .\]
The difference between the total and model Hamiltonians is just the time-varying potential and
\[ i \hbar \frac{d}{dt} \left| \lambda(t) \right> = e^{i H_0(t – t_0)/\hbar} V(t) e^{-i H_0(t – t_0)/\hbar} \left| \lambda(t) \right> \equiv V_I(t) \left| \lambda(t) \right> \; , \]
where $$V_I(t) = U_0(t_0,t) V(t) U_0(t,t_0)$$. The ‘I’ subscript indicates that the potential is now specified in the interaction picture. The time evolution of the state $$\left| \lambda(t) \right>$$ leads immediately to the equation of motion
\[ i \hbar \frac{d}{dt} U_I(t,t_0) = V_I(t) U_I(t,t_0) \]
for the evolution operator $$U_I$$ in the interaction picture. The fact that $$U_I$$ evolves only under the action of $$V_I$$ justifies the name ‘interaction picture’.

What to make of the forward and backward propagation in this definition of $$V_I$$? A meaningful interpretation can be made mining the $$U_I$$’s equation of motion as follows.

The formal solution of the equation of motion is
\[ U_I(t,t_0) = Id – \frac{i}{\hbar} \int_{t_0}^t V_I(t’) U(t’,t_0) dt’ \]
but the time dependence of $$V_I$$ means that the iterated solution
\[ U_I(t,t_0) = Id + \sum_{n=1}^{\infty} \left( \frac{-i}{\hbar} \right)^n \int_{t_0}^{t} dt_1 \, \int_{t_0}^{t_1} dt_2 \, … \\ \int_{t_0}^{t_{n-1}} dt_n \, V_I(t_1) V_I(t_2)…V_I(t_n) \]
from case 3 in Part 1 is the only one available.
To understand what’s happening physically, let’s keep terms in this solution only up to $$n=1$$. Doing so yields
\[ U_I(t,t_0) = Id -\frac{i}{\hbar} \int_{t_0}^t \, dt_1 V_I(t_1) \]
or, expanding $$V_I$$ by its definition,
\[ U_I(t,t_0) = Id – \frac{i}{\hbar} \int_{t_0}^t \, dt_1 U_0(t_0,t_1) V(t_1) U(t_1,t_0) \; . \]

From the relationships between $$\left| \psi \right>$$ and $$\left| \lambda \right>$$ we have
\[ \left| \psi(t) \right> = U_0(t,t_0) \left| \lambda(t) \right> = U_0(t,t_0) U_I(t,t_0) \left| \lambda(t_0)\right> \\ = U_0(t,t_0) U_I(t,t_0) \left| \psi(t_0) \right> \]
from which we conclude

Pre-multiplying by the model Hamiltonian’s evolution operator $$U_0$$ gives
\[ U(t,t_0) = U_0(t,t_0) – \frac{i}{\hbar} \int_{t_0}^t \, dt_1 \left( U_0(t,t_0) U_0(t_0,t_1) V(t_1) U(t_1,t_0) \right) \; , \]
which simplifies using the composition property of the evolution operators as
\[ U(t,t_0) = U_0(t,t_0) – \frac{i}{\hbar} \int_{t_0}^t \, dt_1 U_0(t,t_1) V(t_1) U(t_1,t_0) \; .\]
This first-order form for the full evolution operator suggests that its action on a state can be thought of as comprised of two parts. The first part corresponds to the evolution of the state under the action of the model Hamiltonian over the entire time span from $$t_0$$ to $$t$$. The second part corresponds to the evolution of the state by $$U_0$$ from $$t_0$$ to $$t_1$$ at which point the state’s motion is perturbed by $$V(t)$$ and then the state merrily goes on its way under $$U_0$$ from $$t_1$$ to $$t$$. In order to get the correct answer to first order, all intermediate times at which this perturbative interaction can occur must be included. A visual way of representing this description is given by the following figure

where the thick double line represents the full evolution operator $$U(t,t_0)$$, the thin single line represents the evolution operator $$U_0$$ and the circles represent the interaction with the potential $$V(t)$$ that can happen at any intermediate time. This interpretation can be carried out to any order in the expansion, with two interaction events (two circles) for $$n=2$$, three interaction events (three circles) for $$n=3$$, and so on.

The formal solution of $$U_I$$ can also be manipulated in the same fashion by pre-multiplying by $$U_0$$ to get
\[ U(t,t_0) = U_0(t,t_0) \\ – \frac{i}{\hbar} \int_{t_0}^t \, dt’ U_0(t,t_0) U_0(t_0,t’) V(t’) \; \; U_0(t’,t_0) U_I(t’,t_0) \]
which simplifies to
\[ U(t,t_0) = U_0(t,t_0) – \frac{i}{\hbar} \int_{t_0}^t \, dt’ U_0(t,t’) V(t’) U(t’,t_0) \; . \]
Projecting this equation onto the position basis using $$\left< \vec r \right|$$, $$\left| \vec r_0 \right>$$ and the closure relation $$\int d^3r’ \left| \vec r’ \right>\left< \vec r’\right|$$ for all intermediate positions gives a relationship for the forward-time propagator (Greens Function) of
\[ K^+(\vec r, t; \vec r_0, t_0) = \int_{t_0}^{t} \, dt’ \int \, d^3r’ \\ K^+_0(\vec r, t; \vec r’, t’) V(\vec r’, t’) \; \; K^+(\vec r’, t’; \vec r_0, t_0) \; \]
(compare, e.g., with equation (36.18) in Schiff). This type of analysis leads to the famous Feynman diagrams.

In the last post, I presented a plausibility argument for the Feynman path integral. Central to this argument is the identification of a relation between the transition amplitude $$\left< \vec r, t \right. \left| \vec r_0, t_0 \right>$$ and the classical action given by
\[ \left< \vec r, t \right. \left| \vec r_0, t_0\right> \sim N e^{\frac{i S}{\hbar}} \;. \]
However, because of the composition property of the quantum propagator, we were forced into evaluating the action not just for the classical path but also for all possible paths in that obey causality (i.e. were time ordered).

In this post I will evaluate the closed form for the free particle propagator and will show how to get the same result from the path integral. Along the way, it will also be noted that the same result obtains when only the classical path and action are used. This strange property holds for a variety of systems more complex than the free particle as is proven in Shankar’s book. My presentation here follows his discussion in Chapter 8.

To truly appreciate the pros and cons of working with the path integral, let’s start by first deriving the quantum propagator for the free particle using a momentum space representation. To keep the computations clearer, I will work only in one dimension. The Hamiltonian for a free particle is given in the momentum representation by
\[ H = \frac{p^2}{2m} \]
and in the position representation by
\[ H = -\frac{\hbar^2}{2m} \frac{\partial^2}{\partial x^2} \; .\]
Since the Hamiltonian is purely a function of momentum and is an algebraic function in the momentum representation it is a easier to work with than if it were expressed in the position representation.

Because the momentum operator $$\hat P$$ commutes with the Hamiltonian, the momentum eigenkets for a natural diagonal basis for the Hamiltonian with the energy being given by
\[ E(p) = p^2/2m \; \]
This basis is clearly doubly degenerate for each given value of $$E_p$$ with a right-going momentum $$p_R = +\sqrt{2mE_p}$$ and a left-going momentum $$p_L = -\sqrt{2mE_p}$$ both having the same energy.

The Schrodinger equation in the momentum representation is
\[ \frac{p^2}{2m} \psi(p,t) = i \hbar \partial_t \psi(p,t) \; , \]
which has easily-obtained solutions
\[ \psi(p,t) = e^{-\frac{i}{\hbar} \frac{p^2}{2m} (t-t_0)} \; .\]
The quantum evolution operator can now be expressed in the momentum basis as
\[ U(t_f,t_0) = \int_{-\infty}^{\infty} dp \, \left| p \right> \left< p \right| e^{-\frac{i}{\hbar} \frac{p^2}{2m} (t-t_0)} \; \] By sandwiching the evolution operator between two position eigenkets \[ K^+(x_f,t_f;x_0,t_0) = \left< x_f \right| U(t_f,t_0) \left| x_0 \right> \theta(t_f-t_0)\]
we arrive at the expression for the forward-time propagator
\[ K^+(x_f,t_f;x_0,t_0) = \int_{-\infty}^{\infty} dp \, \left< x_f \right. \left| p \right> \left< p \right. \left| x_0 \right> e^{-\frac{i}{\hbar} \frac{p^2}{2m} (t_f-t_0)} \theta(t_f-t_0)\; .\]
In the remaining computations, it will be understood that $$t_f > t_0$$ and so I will drop the explicit reference to the Heaviside function. This equation can be can be evaluated by using
\[ \left< p | x \right> = \frac{1}{\sqrt{2 \pi \hbar}} e^{-\frac{ipx}{\hbar} } \]to give
\[ K^+(x_f,t_f;x_0,t_0) = \frac{1}{2 \pi \hbar} \int_{-\infty}^{\infty} dp \, e^{\frac{i}{\hbar}p(x_f-x_0)} e^{-\frac{i}{\hbar} \frac{p^2}{2m} (t_f-t_0)}\; .\]

The integral for the propagator can be conveniently written as
\[ K^+(x_f,t_f;x_0,t_0) = \frac{1}{2 \pi \hbar} \int_{-\infty}^{\infty} dp \, e^{-a p^2 + b p} \; , \]
where
\[ a = \frac{i (t_f – t_0)}{2 m \hbar} \]
and
\[ b = \frac{i(x_f – x_0)}{\hbar} \; .\]

Using the standard Gaussian integral
\[ \int_{-\infty}^{\infty} dx \, e^{-ax^2+bx} = e^{b^2/2a} \sqrt{\frac{\pi}{a}} \; ,\]
we arrive at the exact answer for the free-particle, forward-time quantum propagator
\[ K^+(x_f,t_f;x_0,t_0) = \sqrt{ \frac{m}{2\pi i\hbar(t_f-t_0)} } e^{\frac{i m}{2\hbar}\frac{ (x_f-x_0)^2}{(t_f-t_0)}} \; .\]

Now we turn to performing the same computation using the path integral approach. The first step is to express the classical action
as a function of the path. The Lagrangian only consists of the kinetic energy and so
\[ S = \int_{t_0}^{t_f} L[x(t)] dt = \int_{t_0}^{t_f} \frac{1}{2} m {\dot x} ^2 \; .\]
The basic idea of the path integral is to look at the quantum evolution across many small time steps so that each step can be handled more easily. In keeping with this idea, the action integral can be approximated as a sum by the expression
\[ S = \sum_{i=0}^{N-1} \frac{m}{2} \left( \frac{x_{i+1} – x_{i}}{\epsilon} \right)^2 \epsilon \; ,\]
where $$\epsilon$$ is the time step. The forward-time propagator is now written as
\[ K^+(x_f,t_f;x_0,t_0) = \lim_{N\rightarrow \infty, \epsilon \rightarrow 0} Q \int_{-\infty}^{\infty} dx_1 \int_{-\infty}^{\infty} dx_2 … \\ \exp \left[ \frac{i m}{2 \hbar} \sum_{i=0}^{N-1} \frac{(x_{i+1} – x_{i})^2}{\epsilon} \right] \; ,\]
where $$Q$$ is a normalization that will have to be determined at the end. The form of the action gives us hope that these integrals can be evaluated, since the term $$x_{i+1}-x_i$$ connects the positions on only two time slices. For notational convenience we’ll define an intermediate set of integrals
\[ I = \lim_{N\rightarrow\infty} Q \int_{-\infty}^{\infty} dx_1…dx_n \\ \exp \left[ i q \left( (x_N-x_{N-1})^2 + … + (x_2 – x_1)^2 + (x_1 – x_0)^2 \right) \right] \; \]
with $$q = \frac{m}{2 \hbar \epsilon}$$.

To start, let’s work on the $$x_1$$ integral. Since it only involves $$x_0$$ and $$x_2$$ this amounts to solving
\[ I_1 = \int_{-\infty}^{\infty} dx_1 \exp \left\{ i q \left[ 2 x_1^2 – 2(x_2 + x_0) x_1 + (x_2^2 + x_0^2) \right] \right\} \; ,\]
which can be done using essentially the same Gaussian integral as above
\[ \int_{-\infty}^{\infty} e^{-ax^2 + bx + c} = exp(b^2/4a +c) \sqrt{\frac{\pi}{a}} \; .\]
This results in
\[ I_1 = \frac{1}{\sqrt{2}} \left( \frac{i\pi}{q} \right)^{1/2} e^{i q \frac{(x_2-x_0)}{2}} \; .\]
Now the next integral to solve is
\[ I_2 = \int_{-\infty}^{\infty} dx_2 \exp \left\{ i q \left[ (x_3-x_2)^2 + (x_2-x_0)^2/2 \right] \right\} \; .\]
Rather than go through this in detail, I wrote some Maxima code to carry these integrals out

calc_int(integrand,ivar) := block([f,a,b,c],
                                  f : integrand,
                                  a : coeff(expand(f),ivar^2),
                                  b : coeff(expand(f),ivar),
                                  c : ratsimp(f-a*ivar^2-b*ivar),
                                  a : -1*a,
                                  sqrt(%pi/a)*exp(factor(ratsimp(b^2/(4*a) + c))

and the results for up through $$I_4$$ are
\[ I_2 = \frac{1}{\sqrt{3}} \left( \frac{i\pi}{q} \right)^{2/2} e^{i q \frac{(x_3-x_0)}{3}} \; , \]
\[ I_3 = \frac{1}{\sqrt{4}} \left( \frac{i\pi}{q} \right)^{3/2} e^{i q \frac{(x_4-x_0)}{4}} \; ,\]
and
\[ I_4 = \frac{1}{\sqrt{5}} \left( \frac{i\pi}{q} \right)^{4/2} e^{i q \frac{(x_5-x_0)}{5}} \; \]
yielding the result for $$N$$
\[ I = Q \frac{1}{\sqrt{N}} \left( \frac{i\pi}{q} \right)^{\frac{N-1}{2}} e^{i q \frac{(x_N-x_0)}{N}} \; .\]
With all the factors in $$q$$ now fully restored, we get
\[ I = \frac{Q}{\sqrt{N}} \left( \frac{2 \pi i \hbar \epsilon}{m} \right)^{\frac{N-1}{2}} e^{\frac{i m (x_N-x_0)^2}{2 \hbar N \epsilon}} \; .\]
Setting
\[ Q = \left( \frac{m}{2 \pi i \hbar \epsilon} \right)^{\frac{N}{2}} \]
gives
\[ I = \left( \frac{m}{2 \pi i \hbar N \epsilon} \right) e^{\frac{i m (x_N-x_0)^2}{2 \hbar N \epsilon}} \; .\]
Taking the limit as $$N \rightarrow \infty$$, $$\epsilon \rightarrow 0$$, and $$N \epsilon = (t_f – t_0)$$ yields
\[ K^+(x_f,t_f;x_0,t_0) = \sqrt{ \frac{m}{2\pi\hbar i (t_f-t_0)} } e^{i m (x_f-x_0)^2/2\hbar (t_f-t_0)} \; ,\]
which is the exact answer that was obtained above.

While this was a lot more work than the momentum-representation path, it is interesting to note that Shankar proves that
any potential with the form
\[ V = a + bx + c x^2 + d \dot x + e x \dot x \]
yields immediately the forward-time propagator
\[K^+(x_f,t_f;x_0,t_0) = e^{i S_{cl}/h} Q(t) \]
where $$S_{cl}$$ is the classical action and $$Q(t)$$ is a function solely of time that generally cannot be determined. Shankar shows, in the case of a free particle, that
\[ S_{cl} = \int_{t_0}^{t_f} L[x(t)] dt = \frac{m v_{av}^2}{2} (t_f-t_0) = \frac{m}{2} \frac{(x_f-x_0)^2}{t_f-t_0} \]
yielding
\[ K^+(x_f,t_f;x_0,t_0) = Q(t) \exp \left[ \frac{i m (x_f – x_0)^2}{2 \hbar (t_f – t_0)} \right] \; , \]
where $$Q(t)$$ can be determined from the requirement that $$K^+(x_f,t_f;x_0,t_0) = \delta(x_f – x_0)$$ when $$t_f = t_0$$. The applicable identity is
\[ \delta(x_f – x_0) = \lim_{\sigma \rightarrow 0} \frac{1}{\sqrt{\pi \sigma^2}} \exp \left[ -\frac{(x_f – x_0)^2}{\sigma^2} \right] \]
from which we immediately get
\[ Q(t) = \sqrt{ \frac{m}{2\pi \hbar i (t_f-t_0)}} \; .\]
These results mean that a host of practical problems that have intractable propagators in any given representation (due to the need to find the eigenfunctions and then plug them into a power series representing the exponential) can now be calculated with relative ease.