A.12 Heisenberg picture

This book follows the formulation of quantum mechanics as developed by Schrö­din­ger. However, there is another, earlier, formulation due to Heisenberg. This subsection gives a brief description so that you are aware of it when you run into it in literature.

In the Schrö­din­ger picture, physical observables like position and momentum are represented by time-independent operators. The time dependence is in the wave function. This is somewhat counterintuitive because classically position and momentum are time dependent quantities. The Heisenberg picture removes the time dependence from the wave function and absorbs it into the operator.

To see how that works out, consider first the general form of the wave function. It can be written as

\begin{displaymath}
\Psi(\ldots;t) = e^{-{\rm i}H t/\hbar} \Psi(\ldots;0)
\end{displaymath} (A.37)

where the exponential of an operator is defined through its Taylor series:
\begin{displaymath}
e^{-{\rm i}H t/\hbar} = 1 - {\rm i}\frac{t}{\hbar} H
- \frac{t^2}{2!\hbar^2} H^2 + \ldots
\end{displaymath} (A.38)

(To check the above expression for the wave function, take the initial wave function to be any energy eigenfunction of energy $E$. You get the correct $e^{-{{\rm i}}Et/\hbar}$ time dependence, chapter 7.1.2. Every $H$ becomes an $E$. And if the expression works for any eigenfunction, it works for all their combinations too. That means that it works for any wave function, because the eigenfunctions are complete. To be sure, the above form of the wave function applies only if the Hamiltonian is independent of time. Even if it is not, the transformation from the initial wave function $\Psi(\ldots;0)$ to a later one $\Psi(\ldots;t)$ still remains a “unitary” one; one that keeps the wave function normalized. But then you will need to use the Schrö­din­ger equation directly to figure out the time dependence.)

Now consider an arbitrary Schrö­din­ger operator $\widehat{A}$. The physical effects of the operator can be characterized by inner products, as in

\begin{displaymath}
\langle \Psi_1(\ldots;t)\vert\widehat A \Psi_2(\ldots;t)\rangle
\end{displaymath} (A.39)

Such a dot product tells you what amount of a wave function $\Psi_1$ is produced by applying the operator on a wave function $\Psi_2$. Knowing these inner products for all wave functions is equivalent to knowing the operator.

If the time-dependent exponentials are now peeled off $\Psi_1$ and $\Psi_2$ and absorbed into the operator, you get the time-dependent Heisenberg operator

\begin{displaymath}
\widetilde A \equiv e^{{\rm i}H t/\hbar} \widehat A e^{-{\rm i}H t/\hbar}
\end{displaymath} (A.40)

Heisenberg operators will be indicated with a tilde instead of a hat. Note that the argument of the first exponential changed sign because it was taken to the other side of the inner product.

The operator $\widetilde{A}$ depends on time. To see how it evolves, differentiate the product with respect to time:

\begin{displaymath}
\frac{{\rm d}\widetilde A}{{\rm d}t} =
\frac{{\rm i}}{\h...
...bar} \widehat A e^{-{\rm i}H t/\hbar} \frac{{\rm i}}{\hbar} H
\end{displaymath}

The first and third terms can be recognized as a multiple of the commutator of $H$ and $\widetilde{A}$, while the middle term is the Heisenberg version of the time derivative of $\widehat{A}$, in case $\widehat{A}$ does happen to depend on time. So the evolution equation for the Heisenberg operator becomes
\begin{displaymath}
\frac{{\rm d}\widetilde A}{{\rm d}t} = \frac{{\rm i}}{\hba...
... t/\hbar}
\left[H,\widehat A\right] e^{-{\rm i}H t/\hbar} %
\end{displaymath} (A.41)

(Note that there is no difference between the Hamiltonians $\widehat{H}$ and $\widetilde{H}$ because $H$ commutes with itself, hence with its exponentials.)

For example, consider the Schrö­din­ger ${\widehat x}$ position and ${\widehat p}_x$ linear momentum operators of a particle. These do not depend on time. Using the commutators as figured out in chapter 7.2.1, the corresponding Heisenberg operators evolve as:

\begin{displaymath}
\frac{{\rm d}\widetilde x}{{\rm d}t} = \frac{1}{m} \wideti...
..._x}{{\rm d}t} =
- \widetilde{\frac{\partial V}{\partial x}}
\end{displaymath}

Those have the exact same form as the equations for the classical position and momentum of the particle.

In fact, the equivalent of the general equation (A.41) is also found in classical physics: it is derived in advanced mechanics, with the so-called “Poisson bracket” taking the place of the commutator. As a simple example, consider one-di­men­sion­al motion of a particle. Any variable $a$ that depends on the position and linear momentum of the particle, and maybe also explicitly on time, has a time derivative given by

\begin{displaymath}
\frac{{\rm d}a}{{\rm d}t} =
\frac{\partial a}{\partial x...
...\frac{{\rm d}p_x}{{\rm d}t} +
\frac{\partial a}{\partial t}
\end{displaymath}

according to the total differential of calculus. And from the classical Hamiltonian

\begin{displaymath}
H = \frac{p_x^2}{2m} + V
\end{displaymath}

it is seen that the time derivatives of position and momentum obey the classical “Hamiltonian dynamics”

\begin{displaymath}
\frac{{\rm d}x}{{\rm d}t} = \frac{\partial H}{\partial p_x...
...\frac{{\rm d}p_x}{{\rm d}t} = - \frac{\partial H}{\partial x}
\end{displaymath}

Substituting this into the time derivative of $a$ gives

\begin{displaymath}
\frac{{\rm d}a}{{\rm d}t} =
\frac{\partial a}{\partial x...
...rac{\partial H}{\partial x} +
\frac{\partial a}{\partial t}
\end{displaymath}

The first two terms in the right hand side are by definition minus the Poisson bracket $\{H,a\}_{\rm {P}}$, so

\begin{displaymath}
\frac{{\rm d}a}{{\rm d}t} = - \{H,a\}_{\rm {P}} + \frac{\p...
...\frac{\partial a}{\partial x} \frac{\partial H}{\partial p_x}
\end{displaymath}

Note that the Poisson bracket, like the commutator, is antisymmetric under exchange of $H$ and $a$. Apparently, formally identifying the Poisson bracket with the commutator divided by ${\rm i}\hbar$ brings you from classical mechanics to Heisenberg’s quantum mechanics.

More generally, the classical Hamiltonian can depend on multiple and non-Cartesian coordinates, generically called “generalized coordinates.” In that case, in the Poisson bracket you must sum over all generalized coordinates and their associated so-called “canonical” momenta. For a Cartesian position coordinate, the canonical momentum is the corresponding linear momentum. For an angular coordinate, it is the corresponding angular momentum. In general, using the so-called Lagrangian formulation usually covered in an engineering education, and otherwise found in addendum {A.1}, the canonical momentum is the derivative of the Lagrangian with respect to the time derivative of the coordinate.

The bottom line is that the Heisenberg equations are usually not easy to solve unless you return to the Schrö­din­ger picture by peeling off the time dependence. In relativistic applications however, time joins space as an additional coordinate, and the Heisenberg picture becomes more helpful. It can also make it easier to identify the correspondence between classical equations and the corresponding quantum operators.


Key Points
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In the Heisenberg picture, operators evolve in time just like their physical variables do in classical physics.