Quantum Evolution – Part 5
This post is a prelude to the final set of posts that transform the evolution/propagation machinery that has been developed into the spacetime picture needed to appreciate Feynman’s work and to act as a bridge to quantum field theory. The subject of this post, the various pictures in quantum mechanics (Schrodinger and Heisenberg) is one that I find particularly confusing due to what I would call an overly haphazard development in most of the textbooks.
As I’ve discussed in my post on the lack of coverage the Helmholtz theorem receives in text books, one of the greatest disservices that is visited on the student is the teaching of a concept that must then be untaught. No presentation seems to me to be as fraught with this difficulty as the discussion associated with the various quantum pictures in terms of fixed and variable states, basis states, and operators. It smacks of the similar confusion that is often engendered between active and passive transformations and rates of change in fixed and rotating frames, but it is compounded by a larger number of objects and a corresponding lack of attention to detail by most authors.
To give a tangible example, consider the coverage of quantum dynamics and evolution in Chapter 2 of Modern Quantum Mechanics by J.J. Sakurai. Sakurai goes to great pains earlier in the chapter (pages 72-3) to distinguish the three cases that must be considered when constructing the propagator. He then promptly drops the most general case where the Hamiltonian is time-dependent and does not commute with itself at different times in his treatment of the Schrodinger and Heisenberg pictures. Even worse, he explicitly steers the student away from the correct general result when he says (page 83)
Because $$H$$ was originally introduced in the Schrodinger picture, we may be tempted to define
\[ H_H = U^{\dagger} H U \]
in accordance with [the definition of operators in the Heisenberg picture]. But in elementary applications where $$U$$ is given by [$$exp(-i H t/ \hbar)$$], $$U$$ and $$H$$ obviously commute; as a result
\[ U^{\dagger} H U = H \]
The use of the word ‘tempted’ makes it sound like one is making a mistake with that first definition, when that first definition is always correct, and it is our use of the second which is a temptation that should be carefully indulged. The similar kind of sloppiness holds true for the works by Shankar and Schiff. Only Cohen-Tannoudji et. al. cover the materially carefully but unfortunately too briefly to really help (or even to be understandable if you don’t know what details to watch).
So what I am presenting here is the most careful and comprehensive way to treat the development of the these two pictures that I know. I’ve patterned it as an amalgam of Schiff in its basic attack and Cohen-Tannoudji in its care for the details joined with my own approach in explaining the physics and in providing a clear notation.
The starting point is the identification of the Schrodinger picture as the one in which the time evolution of the state is given by the familiar equation
\[ i \hbar \frac{d}{dt} \left| \psi(t) \right> = H \left| \psi(t) \right> \; . \]
A point on notation before proceeding. Where needed, an object that is in the Schrodinger picture will be decorated with the subscript ‘S’ and, likewise, an object in the Heisenberg picture will always have an ‘H’ subscript. An object with no decoration is understood to be in the Schrodinger picture.
Start with a Schrodinger picture operator
\[ \Omega_S = \Omega_S (t) \]
that generally has a time dependence, which, for notational simplicity, will be suppressed in what follows. A convenient physical picture is to imagine that $$\Omega_S$$ is a time dependent measurement, like what would result from a Stern-Gerlach apparatus that is rotating uniformly in space as a function of time.
At any given time, imagine the state to be given by $$\left| \psi(t) \right>$$ and ask what overlap the state has with the state $$\left| \lambda(t) \right>$$ after being subjected to the operation of $$\Omega_S$$. The expected overlap (or projection) is defined as
\[ \left< \Omega_S \right>_{\lambda \psi} \equiv \left< \lambda(t) | \Omega_S |\psi(t) \right> \; . \]
Now ask how this expected overlap changes as a function of time, remembering that both the operator and the state are changing. Taking the appropriate time derivative of $$\left< \Omega_S \right>_{\lambda \psi}$$ and expanding yields
\[ \frac{d}{dt} \left< \Omega_S \right>_{\lambda \psi} = \left[ \frac{d}{dt} \left< \lambda (t) \right| \right] \Omega_S \left| \psi(t) \right> + \left< \lambda(t) \left| \frac{\partial \Omega_S}{\partial t} \right| \psi(t) \right> \\ + \left< \lambda(t) \right| \Omega_S \left[ \frac{d}{dt} \left| \psi(t) \right> \right] \; .\]
Each state obeys the time-dependent Schrodinger equation
\[ i \hbar \frac{d}{dt} \left| \psi(t) \right> = H \left| \psi(t) \right> \]
and
\[ – i \hbar \frac{d}{dt} \left< \lambda(t) \right| = \left< \lambda(t) \right| H \; , \]
where the fact that the Hamiltonian is Hermitian ($$H^{\dagger} = H$$) is used for the dual equation involving the bra $$\left< \lambda(t) \right|$$.
The time derivatives can be eliminated in favor of the multiplication of the Hamiltonian. Substituting these results in and grouping terms yields
Note that I’ve broken with tradition by not denoting the first term as $$\left< \frac{\partial \Omega_S}{\partial t} \right>_{\lambda\psi}$$. The partial derivative notation is meant to motivate the transition from classical to quantum mechanics (the evolution of a classical function in terms of Poisson brackets) and was used a lot in the origins of the subject. However, there is nothing partial about the time dependence of the operator $$\Omega_s$$ since it only depends on time.
This expression is not particularly satisfactory since the arbitrary state vectors $$\left| \psi (t) \right>$$ and $$\left| \lambda (t) \right>$$ are still present. There is a way to push the time dependence onto the operators completely by going to the Heisenberg picture (sometimes it is said that this is a frame that co-moves with the state vectors themselves).
Since each state obeys the time-dependent Schrodinger equation, its time evolution can be written as
\[ \left< \lambda(t) \right| = \left< \lambda(t_0) \right| U^{\dagger}(t,t_0) \] and \[ \left| \psi(t) \right> = U(t,t_0) \left| \psi(t_0) \right> \; .\]
Substitution of the right-hand side of these equations expresses the expected overlap in terms of the states at the fixed time $$t_0$$
\[ \frac{d}{dt} \left< \Omega_S \right>_{\lambda \psi} = \frac{d}{dt} \left< \lambda(t_0) \left| U^{\dagger}(t,t_0) \Omega_S U(t,t_0) \right| \psi(t_0) \right> \]
The time derivative now passes into the expectation to hit the operators directly
\[\frac{d}{dt} \left< \lambda(t_0) \left| U^{\dagger}(t,t_0) \Omega_S U(t,t_0) \right| \psi(t_0) \right> \\ = \left< \lambda(t_0) \left| \frac{d}{dt}\left( U^{\dagger}(t,t_0) \Omega_S U(t,t_0)\right) \right| \psi(t_0) \right> \; ,\]
and, as a result of the arbitrariness of the state vectors, this middle piece can be liberated and subsequently simplified by expanding using the product rule
\[ \frac{d}{dt}\left( U^{\dagger}(t,t_0) \Omega_S U(t,t_0)\right) = \left( \frac{d}{dt} U^{\dagger}(t,t_0) \right) \Omega_S U(t,t_0) \\ + U^{\dagger}(t,t_0) \left( \frac{d}{dt} \Omega_S \right) U(t,t_0) + U^{\dagger}(t,t_0) \Omega_S \frac{d}{dt}\left( U(t,t_0)\right) \; .\]
The time derivatives of the evolution operators, which are given by analogous formulas to the state propagation
\[ \frac{d}{dt}U(t,t_0) = -\frac{1}{i \hbar} H U(t,t_0) \]
and
\[ \frac{d}{dt}U^{\dagger}(t,t_0) = \frac{1}{i \hbar} U^{\dagger}(t,t_0) H \; ,\]
produce a further simplification to
\[ \frac{d}{dt}\left( U^{\dagger}(t,t_0) \Omega_S U(t,t_0)\right) = U^{\dagger}(t,t_0) \left( \frac{d}{dt} \Omega_S \right) U(t,t_0) \\ + \frac{1}{i \hbar} U^{\dagger}(t,t_0) [\Omega_S,H] U(t,t_0) \]
It is attractive to define the operator $$\Omega$$ in the Heisenberg picture through the identification of\[ \Omega_H \equiv U^{\dagger}(t,t_0) \Omega_S U(t,t_0) \]
and somewhat awkward definition
\[ \left( \frac{d}{dt} \Omega_S \right)_H \equiv U^{\dagger}(t,t_0) \left( \frac{d}{dt} \Omega_S \right) U(t,t_0) \; ,\]
where I am favoring the careful notation of Cohen-Tannoudji.
These identifications produce the expression
\[ \frac{d \Omega_H}{d t} = \left( \frac{d \Omega_S}{d t} \right)_H + \frac{1}{i \hbar} U^{\dagger}(t,t_0) [\Omega_S,H] U(t,t_0) \]
that looks like it wants to become the classical equation for the total time derivative of a function expressed in terms of the Poisson bracket
\[ \frac{d}{dt} F = \frac{\partial}{\partial t} F + [F,H] \]
where the brackets here are of the Poisson, not commutator, variety.
A cleaner identification can be made between classical and quantum mechanics as follows. Since the time evolution arguments are understood to be from $$t_0$$ to $$t$$ whenever a propagator $$U$$ is encountered, they will be suppressed.
First expand the commutator
\[ U^{\dagger}[\Omega_S,H] U = U^{\dagger} H \Omega_S U – U^{\dagger} \Omega_S H U \]
and then insert if $$U^{\dagger} U = Id$$ in strategic places to get
\[U^{\dagger} H U U^{\dagger} \Omega_S U – U^{\dagger} \Omega_S U U^{\dagger} H U = U^{\dagger} H U \Omega_H – \Omega_H U^{\dagger} H U \; . \]
Finally identify the Hamiltonian in the Heisenberg picture as
\[ H_H = U^{\dagger} H U \; \]
and rewrite the equation as (see also equation (8) in Complement $$G_{III}$$ of Cohen-Tannoudji)
Most authors are not clear in the statements they make about the differences between the Hamiltonian in the two pictures, tending to confuse the general rule that the two Hamiltonians differ (as they should since this movement from the Schrodinger to the Heisenberg picture is a canonical transformation) with the special case when they do. This special case occurs in the usual textbook treatment of a time-independent Hamiltonian, where the propagator is given by
\[ U(t,t_0) = e^{-i H (t-t_0)/ \hbar } \]
and in this case $$H_H = H$$.
It also follows that, in this case, if $$\Omega_S$$ does not depend on time and commutes with $$H$$ then it is a conserved quantity and its corresponding operator in the Heisenberg picture is as well.