To briefly review, we've gone through three concrete problems in the last couple of lectures, and in each case we've used a somewhat different approach to solve for the behavior:
There's a larger point behind this list of examples, which is that our "quantum toolkit" of problem-solving methods contains many approaches: we can often use more than one method for a given problem, but often it's easiest to proceed using one of them. (We could have used operator algebra for Larmor precession, for example, by summing the power series to get \( \hat{U}(t) \).)
Over the rest of the semester, we'll be making use of all three approaches depending on the problem. Knowing which method to apply to a specific problem is an art - something you have to get a feel for by solving problems and seeing examples. This is, of course, not new in physics: in classical mechanics you already know that you can apply Newton's laws, or conservation of energy, or the Lagrangian, or the Hamiltonian, and the best choice will vary by what system you're studying and what question you're asking. But if you're used to quantum mechanics as wave mechanics, then you'll have to adjust to the new methods being available.
Now, let's talk more generally about operator algebra and time evolution. So far, we have studied time evolution in the Schrödinger picture, where state kets evolve according to the Schrödinger equation
\[ \begin{aligned} i \hbar \frac{d}{dt} \ket{\psi(t)} = \hat{H} \ket{\psi(t)}, \end{aligned} \]
while operators (and thus basis kets) are time-independent. As we saw, when \( \hat{H} \) is time-independent, we can formally integrate this equation to obtain
\[ \begin{aligned} \ket{\psi(t)} = e^{-i \hat{H} t/\hbar} \ket{\psi(0)} \equiv \hat{U}(t) \ket{\psi(0)}, \end{aligned} \]
i.e. time evolution is just the result of a unitary operator \( \hat{U} \) acting on the kets.
What about the more general case? It turns out that time evolution can always be thought of as equivalent to a unitary operator acting on the kets, even when the Hamiltonian is time-dependent. First, suppose that \( \hat{H} \) depends explicitly on time but commutes with itself at different times, e.g. a spin-1/2 particle interacting with a background magnetic field whose direction is fixed but whose magnitude changes,
\[ \begin{aligned} \hat{H} = \frac{e}{mc} \hat{S_z} B_z(t). \end{aligned} \]
Here we can still solve the Schrödinger equation just by formally integrating both sides, but now that \( \hat{H} \) depends on time we end up with an integral in the exponential,
\[ \begin{aligned} \hat{U}(t) = \exp \left[ - \left(\frac{i}{\hbar}\right) \int_0^t dt'\ \hat{H}(t') \right]. \end{aligned} \]
There exist even more complicated cases where the Hamiltonian doesn't even commute with itself at different times. In fact, we just saw such an example; the spin-1/2 particle in a magnetic field which rotates in the \( xy \) plane gives a Hamiltonian such that \( [\hat{H}(t), \hat{H}(t')] \neq 0 \). There is, nevertheless, still a formal solution known as the Dyson series,
\[ \begin{aligned} \hat{U}(t) = 1 + \sum_{n=1}^\infty \left( \frac{-i}{\hbar} \right)^n \int_0^t dt_1 \int_0^{t_1} dt_2 ... \int_0^{t_{n-1}} dt_n \hat{H}(t_1) \hat{H}(t_2)...\hat{H}(t_n). \end{aligned} \]
(This is a good time to appreciate the fact that we didn't have to use the formal solution for the two-state system!) It's not self-evident that these more complicated constructions are still unitary, especially the Dyson series, but rest assured that they are. So time evolution is always a unitary transformation acting on the states.
As we observed before, this implies that inner products of state kets are preserved under time evolution:
\[ \begin{aligned} \sprod{\alpha(t)}{\beta(t)} = \bra{\alpha(0)} \hat{U}{}^\dagger(t) \hat{U}(t) \ket{\beta(0)} = \sprod{\alpha(0)}{\beta(0)}. \end{aligned} \]
On the other hand, the matrix elements of a general operator \( \hat{A} \) will be time-dependent, unless \( \hat{A} \) commutes with \( \hat{U} \):
\[ \begin{aligned} \bra{\alpha(t)} \hat{A} \ket{\beta(t)} = \bra{\alpha(0)} \hat{U}{}^\dagger(t) \hat{A} \hat{U}(t) \ket{\beta(0)} \end{aligned} \]
Notice that by definition in the Schrödinger picture, the unitary transformation only affects the states, so the operator \( \hat{A} \) remains unchanged. But now, we can see that we could have equivalently left the state vectors unchanged, and evolved the observable in time, i.e.
\[ \begin{aligned} \bra{\alpha} \hat{A}(t) \ket{\beta} = \bra{\alpha} (\hat{U}{}^\dagger (t) \hat{A}(0) \hat{U}(t)) \ket{\beta}. \end{aligned} \]
This is exactly the same product of states and operators; we get the same answer. But now all of the time dependence has been pushed into the observable. This is the Heisenberg picture of quantum mechanics.
Which picture is better to work in? There's no definitive answer; the two pictures are useful for answering different questions. In particular, we might guess that the Heisenberg picture would make it easier to connect with classical mechanics; in the classical world, observables themselves (things like position \( \vec{x} \) or angular momentum \( \vec{L} \)) are the things which evolve in time, whereas there's no classical analogue to the state vector.
Let's make our notation explicit. We define the Heisenberg picture observables by
\[ \begin{aligned} \hat{A}{}^{(H)}(t) \equiv \hat{U}{}^\dagger(t) \hat{A}{}^{(S)} \hat{U}(t), \end{aligned} \]
where \( (H) \) and \( (S) \) stand for Heisenberg and Schrödinger pictures, respectively. The two operators are equal at \( t=0 \), by definition; \( \hat{A}^{(S)} = \hat{A}(0) \). On the other hand, in the Heisenberg picture the state vectors are frozen in time,
\[ \begin{aligned} \ket{\alpha(t)}_H = \ket{\alpha(0)} \end{aligned} \]
whereas in the Schrödinger picture we have
\[ \begin{aligned} \ket{\alpha(t)}_S = \hat{U}(t) \ket{\alpha(0)}. \end{aligned} \]
As we've observed, expectation values are the same, no matter what picture we use, as they should be (the choice of picture itself is not physical.)
We can derive an equation of motion for the operators in the Heisenberg picture, starting from the definition above and differentiating:
\[ \begin{aligned} \frac{d\hat{A}{}^{(H)}}{dt} = \frac{\partial \hat{U}{}^\dagger}{\partial t} \hat{A}{}^{(S)} \hat{U} + \hat{U}{}^\dagger \hat{A}{}^{(S)} \frac{\partial \hat{U}}{\partial t} \\ = -\frac{1}{i\hbar} \hat{H} \hat{U}{}^\dagger \hat{A}{}^{(S)} \hat{U} + \frac{1}{i\hbar} \hat{U}{}^\dagger \hat{A}{}^{(S)} \hat{U} \hat{H} \\ \Rightarrow \frac{d \hat{A}{}^{(H)}}{dt} = \frac{1}{i\hbar} [\hat{A}{}^{(H)}, \hat{H}]. \end{aligned} \]
This is the Heisenberg equation of motion, and I've made use of the fact that the unitary operator \( \hat{U} \), which is constructed from \( \hat{H} \), certainly commutes with \( \hat{H} \). We have assumed here that the Schrödinger picture operator is time-independent, but sometimes we want to include explicit time dependence of an operator, e.g. a time-varying external magnetic field. So the complete Heisenberg equation of motion should be written
\[ \begin{aligned} \frac{d\hat{A}{}^{(H)}}{dt} = \frac{1}{i\hbar} [\hat{A}{}^{(H)}, \hat{H}] + \left(\frac{\partial \hat{A}}{dt}\right)^{(H)} \end{aligned} \]
where the last term is related to the Schrödinger picture operator like so:
\[ \begin{aligned} \left(\frac{\partial \hat{A}}{dt}\right)^{(H)} = \hat{U}{}^\dagger \frac{\partial \hat{A}{}^{(S)}(t)}{\partial t} \hat{U}. \end{aligned} \]
Notice that the operator \( \hat{H} \) itself doesn't evolve in time in the Heisenberg picture. (Assuming it has no explicit time dependence, and Heisenberg picture can become very messy if it does!) Likewise, any operators which commute with \( \hat{H} \) are time-independent in the Heisenberg picture.
Actually, this equation requires some explaining, because it immediately contravenes my definition that "operators in the Schrödinger picture are time-independent". The more correct statement is that "operators in the Schrödinger picture do not evolve in time due to the Hamiltonian of the system"; we have to separate out the time-dependence due to the Hamiltonian from explicit time dependence (again, most commonly imposed by the presence of a time-dependent background classical field.)
The Heisenberg equation of motion provides the first of many connections back to classical mechanics. The time evolution of a classical system can be written in the familiar-looking form
\[ \begin{aligned} \frac{dA}{dt} = \{A, H\}_{PB} + \frac{\partial A}{\partial t} \end{aligned} \]
where \( H \) is the Hamiltonian, and the brackets are the Poisson bracket, defined in general as
\[ \begin{aligned} \{f, g\}_{PB} = \sum_i \left( \frac{\partial f}{\partial q_i} \frac{\partial g}{\partial p_i} - \frac{\partial f}{\partial p_i} \frac{\partial g}{\partial q_i}\right). \end{aligned} \]
So the Heisenberg equation of motion can be obtained from the classical one by applying Dirac's quantization rule,
\[ \begin{aligned} \{,\}_{PB} \rightarrow \frac{1}{i\hbar} [,]. \end{aligned} \]
This approach, known as canonical quantization, was one of the early ways to try to understand quantum physics. You should be suspicious about the claim that we can derive quantum mechanics from classical mechanics, and in fact we know that we can't; operators like spin have no classical analogue from which to start. (There are other, more subtle issues; in fact the quantization rule fails even for some observables that do have classical counterparts, if they involve higher powers of \( \hat{x} \) and \( \hat{p} \) for instance.)
Let's have a closer look at some of the parallels between classical mechanics and QM in the Heisenberg picture. First, a useful identity between \( \hat{x} \) and \( \hat{p} \):
\[ \begin{aligned} [\hat{x}, \hat{p}^n] = [\hat{x}, \hat{p} \hat{p}^{n-1}] \\ = \hat{p} [\hat{x}, \hat{p}^{n-1}] + [\hat{x}, \hat{p}] \hat{p}^{n-1} \\ = \hat{p} [\hat{x}, \hat{p}^{n-1}] + i\hbar \hat{p}^{n-1} \\ = \hat{p} \left( \hat{p} [\hat{x}, \hat{p}^{n-2}] + i\hbar \hat{p}^{n-2}\right) + i\hbar \hat{p}^{n-1} \\ = \hat{p}{}^2 [\hat{x}, \hat{p}^{n-2}] + 2i\hbar \hat{p}^{n-1} \\ = (...) \end{aligned} \]
and so on. Expansion of the commutator will terminate at \( [\hat{x}, \hat{p}] = i\hbar \), at which point there will be \( (n-1) \) copies of the \( i\hbar \hat{p}^{n-1} \) term. Thus,
\[ \begin{aligned} [\hat{x}, \hat{p}^n] = ni\hbar \hat{p}^{n-1} \end{aligned} \]
and similarly
\[ \begin{aligned} [\hat{p}, \hat{x}^n] = -ni\hbar \hat{x}^{n-1}. \end{aligned} \]
These can be used with the power-series definition of functions of operators to derive the even more useful identities (now in 3 dimensions)
\[ \begin{aligned} [\hat{x_i}, F(\hat{\vec{p}})] = i \hbar \frac{\partial F}{\partial \hat{p_i}}, \\ [\hat{p_i}, G(\hat{\vec{x}})] = -i \hbar \frac{\partial G}{\partial \hat{x_i}}. \end{aligned} \]
Now we have what we need to return to one of our previous simple examples, the lone particle of mass \( m \):
\[ \begin{aligned} \hat{H} = \frac{\hat{\vec{p}}{}^2}{2m} + V(\hat{\vec{x}}). \end{aligned} \]
Note that I'm not writing any of the \( (H) \) superscripts, since we're working explicitly with the Heisenberg picture there should be no risk of confusion. Before we treat the general case, what does the free particle look like, \( \hat{H}_0 = \hat{\vec{p}}^2/2m \)? First of all, the momentum now commutes with \( \hat{H} \), which means that it is conserved:
\[ \begin{aligned} \frac{d\hat{p_i}}{dt} = \frac{1}{i\hbar} [\hat{p_i}, \hat{H}_0] = 0. \end{aligned} \]
On the other hand, for the position operators we have
\[ \begin{aligned} \frac{d\hat{x_i}}{dt} = \frac{1}{i\hbar} [\hat{x_i}, \hat{H}_0] \ = \frac{1}{2mi\hbar} \left(i\hbar \frac{\partial \hat{H}_0}{\partial \hat{p_i}}\right) \\ = \frac{\hat{p_i}}{m}. \end{aligned} \]
This is exactly the classical definition of the momentum for a free particle, and the trajectory as a function of time looks like a classical trajectory:
\[ \begin{aligned} x_i(t) = x_i(0) + \left( \frac{p_i(0)}{m}\right) t. \end{aligned} \]
One important subtlety that I've glossed over. Now that our operators are functions of time, we have to be careful to specify that the usual set of commutation relations between \( \hat{x} \) and \( \hat{p} \) are now only guaranteed to be true for the original operators at \( t=0 \). Indeed, if we check we find that \( \hat{x}_i(t) \) does not commute with \( \hat{x}_i(0) \):
\[ \begin{aligned} [\hat{x_i}(t), \hat{x_i}(0)] = \left[ \hat{x_i}(0) + \frac{t}{m} \hat{p_i}(0), \hat{x_i}(0) \right] = -\frac{i\hbar t}{m}. \end{aligned} \]
The commutation relations for \( \hat{p}(t) \) are unchanged here, since it doesn't evolve in time. An amusing thing we can do with the commutator of the position operators is apply the uncertainty relation, finding
\[ \begin{aligned} (\Delta x_i(t))^2 (\Delta x_i(0))^2 \geq \frac{\hbar^2 t^2}{4m^2}. \end{aligned} \]
To make sense of this, you could imagine tracking the evolution of e.g. a wave packet initial state: this says that over time, with no potential applied a wave packet will spread out in position space over time. (You can go back and solve for the time evolution of our wave packet using the Schrödinger equation and verify this relation holds!)
Now, we switch back on the potential function \( V(\hat{\vec{x}}) \). This doesn't change our time-evolution equation for the \( \hat{x}_i \), since they commute with the potential. However, for the momentum operators, we now have
\[ \begin{aligned} \frac{d\hat{p_i}}{dt} = \frac{1}{i\hbar} [\hat{p_i}, V(\hat{x})] = -\frac{\partial V}{\partial x_i}. \end{aligned} \]
This should already look familiar, and if we go back and take the time derivative of the \( dx_i/dt \) expression above, we can eliminate the momentum to rewrite it in the more familiar form
\[ \begin{aligned} m \frac{d^2 \hat{\vec{x}}}{dt^2} = - \nabla V(\hat{x}). \end{aligned} \]
This is (the quantum version of) Newton's second law! This derivation depended on the Heisenberg picture, but if we take expectation values then we find a picture-independent statement
\[ \begin{aligned} m \frac{d^2}{dt^2} \ev{\hat{\vec{x}}} = \frac{d}{dt} \ev{\hat{\vec{p}}} = - \ev{\nabla V(\hat{\vec{x}})}. \end{aligned} \]
This relation is known as Ehrenfest's theorem, and was derived by Ehrenfest using wave mechanics (we had the easier path with the Heisenberg picture.) It shows that on average, the center of a quantum wave packet moves exactly like a classical particle.
In the Schrödinger picture, our starting point for any calculation was always with the eigenkets of some operator, defined by the equation
\[ \begin{aligned} \hat{A}{}^{(S)} \ket{a} = a \ket{a}. \end{aligned} \]
The eigenkets \( \ket{a} \) then give us part or all of a basis for our Hilbert space. Since the operator doesn't evolve in time, neither do the basis kets. On the other hand, since the Heisenberg-picture operators do evolve in time, their eigenkets must as well:
\[ \begin{aligned} \hat{A}{}^{(H)}(t) \ket{a,t} = a \ket{a,t}. \end{aligned} \]
(Remember that the eigenvalues are always the same, since a unitary transformation doesn't change the spectrum of an operator!) Expanding out in terms of the operator at time zero,
\[ \begin{aligned} \hat{A}{}^{(H)}(0) \ket{a,0} = a \ket{a,0} \\ \hat{U}{}^\dagger (t) \hat{A}{}^{(H)}(0) (\hat{U}(t) \hat{U}{}^\dagger) \ket{a,0} = a \hat{U}{}^\dagger (t) \ket{a,0} \end{aligned} \]
where I've inserted the identity operator. But now we can see the Heisenberg picture operator at time \( t \) on the left-hand side, and we identify the evolution of the ket,
\[ \begin{aligned} \ket{a,t} = \hat{U}{}^\dagger (t) \ket{a,0}. \end{aligned} \]
This is the opposite direction of how the state evolves in the Schrödinger picture, and in fact the state kets satisfy the Schrödinger equation with the wrong sign,
\[ \begin{aligned} i \hbar \frac{\partial}{\partial t} \ket{a,t} = - \hat{H} \ket{a,t}. \end{aligned} \]
Don't get confused by all of this; all we're doing is grouping things together in a different order!
Next time: a little more on evolution of kets, then the harmonic oscillator again.