7.2 Time Variation of Expectation Values

The time evolution of systems may be found using the Schrödinger equation as described in the previous section. However, that requires the energy eigen­functions to be found. That might not be easy.

For some systems, especially for macroscopic ones, it may be sufficient to figure out the evolution of the expec­tation values. An expec­tation value of a physical quantity is the average of the possible values of that quantity, chapter 4.4. This section will show how expec­tation values may often be found without finding the energy eigen­functions. Some appli­cations will be indicated.

The Schrödinger equation requires that the expec­tation value $\big\langle a\big\rangle$ of any physical quantity $a$ with associated operator $A$ evolves in time as:

$$\fbox{$\displaystyle \frac{{\rm d}\langle a \rangle}{{\r... ...ft\langle \frac{\partial A}{\partial t} \right\rangle $} %$$ (7.4)

A derivation is in {D.35}. The commutator $[H,A]$ of $A$ with the Hamiltonian was defined in chapter 4.5 as $HA-AH$. The final term in (7.4) is usually zero, since most (simple) operators do not explicitly depend on time.

The above evolution equation for expec­tation values does not require the energy eigen­functions, but it does require the commutator.

Note from (7.4) that if an operator $A$ commutes with the Hamiltonian, i.e. $[H,A]$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0, then the expec­tation value of the corre­sponding quantity $a$ will not vary with time. Actually, that is just the start of it. Such a quantity has eigen­functions that are also energy eigen­functions, so it has the same time-conserved statistics as energy, section 7.1.4. The uncertainty, proba­bilities of the individual values, etcetera, do not change with time either for such a variable.

One appli­cation of equation (7.4) is the so-called “virial theorem” that relates the expec­tation potential and kinetic energies of energy eigen­states, {A.17}. For example, it shows that harmonic oscillator states have equal potential and kinetic energies. And that for hydrogen states, the potential energy is minus two times the kinetic energy.

Two other important appli­cations are discussed in the next two subsections.

Key Points

$\begin{picture}(15,5.5)(0,-3) \put(2,0){\makebox(0,0){\scriptsize\bf0}} \put(12... ...\thicklines \put(3,0){\line(1,0){12}}\put(11.5,-2){\line(1,0){3}} \end{picture}$

A relatively simple equation that describes the time evolution of expec­tation values of physical quantities exists. It is fully in terms of expec­tation values.

$\begin{picture}(15,5.5)(0,-3) \put(2,0){\makebox(0,0){\scriptsize\bf0}} \put(12... ...\thicklines \put(3,0){\line(1,0){12}}\put(11.5,-2){\line(1,0){3}} \end{picture}$

Variables which commute with the Hamiltonian have the same time-independent statistics as energy.

$\begin{picture}(15,5.5)(0,-3) \put(2,0){\makebox(0,0){\scriptsize\bf0}} \put(12... ...\thicklines \put(3,0){\line(1,0){12}}\put(11.5,-2){\line(1,0){3}} \end{picture}$

The virial theorem relates the expec­tation kinetic and potential energies for important systems.

7.2.1 Newtonian motion

The purpose of this section is to show that even though Newton's equations do not apply to very small systems, they are correct for macroscopic systems.

The trick is to note that for a macroscopic particle, the position and momentum are very precisely defined. Many unavoidable physical effects, such as incident light, colliding air atoms, earlier history, etcetera, will narrow down position and momentum of a macroscopic particle to great accuracy. Heisenberg's uncertainty relationship says that they must have uncertainties big enough that $\Delta{p}_x\Delta{x}$ $\raisebox{-.5pt}{$\geqslant$}$ $\frac12\hbar$, but $\hbar$ is far too small for that to be noticeable on a macroscopic scale. Normal light changes the momentum of a rocket ship in space only immeasurably little, but it is quite capable of locating it to excellent accuracy.

With little uncertainty in position and momentum, both can be approximated accurately by their expec­tation values. So the evolution of macroscopic systems can be obtained from the evolution equation (7.4) for expec­tation values given in the previous subsection. Just work out the commutator that appears in it.

Consider one-di­mensional motion of a particle in a potential $V(x)$ (the three-di­mensional case goes exactly the same way). The Hamiltonian $H$ is:

$$H = \frac{{\widehat p}_x^2}{2m} + V(x)$$

where ${\widehat p}_x$ is the linear momentum operator and $m$ the mass of the particle.

Now according to evolution equation (7.4), the expec­tation position $\big\langle x\big\rangle$ changes at a rate:

$$\frac{{\rm d}\langle x \rangle}{{\rm d}t} = \left\langle... ...\widehat p}_x^2}{2m} + V(x),{\widehat x}\right] \right\rangle$$ (7.5)

Recalling the properties of the commutator from chapter 4.5, $[V(x),{\widehat x}]$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0, since multi­plication commutes. Further, according to the rules for manipul­ation of products and the canonical commutator

$$[{\widehat p}_x^2,{\widehat x}]= {\widehat p}_x[{\widehat p}_... ...widehat p}_x]{\widehat p}_x = - 2 {\rm i}\hbar {\widehat p}_x$$

So the rate of change of expec­tation position becomes:

$$\frac{{\rm d}\langle x \rangle}{{\rm d}t} = \left\langle \frac{p_x}{m} \right\rangle$$ (7.6)

This is exactly the Newtonian expression for the change in position with time, because Newtonian mechanics defines $p_x$$\raisebox{.5pt}{$/$}$$m$ to be the velocity. However, it is in terms of expec­tation values.

To figure out how the expec­tation value of momentum varies, the commutator $[H,{\widehat p}_x]$ is needed. Now ${\widehat p}_x$ commutes, of course, with itself, but just like it does not commute with ${\widehat x}$, it does not commute with the potential energy $V(x)$. The gener­alized canonical commutator (4.62) says that $[V,{\widehat p}_x]$ equals $\vphantom0\raisebox{1.5pt}{$-$}$$\hbar\partial{V}$$\raisebox{.5pt}{$/$}$${\rm i}\partial{x}$. As a result, the rate of change of the expec­tation value of linear momentum becomes:

$$\frac{{\rm d}\langle p_x \rangle}{{\rm d}t} = \left\langle - \frac{\partial V}{\partial x} \right\rangle$$ (7.7)

This is Newton's second law in terms of expec­tation values: Newtonian mechanics defines the negative derivative of the potential energy to be the force, so the right hand side is the expec­tation value of the force. The left hand side is equivalent to mass times accel­eration.

The fact that the expec­tation values satisfy the Newtonian equations is known as “Ehrenfest's theorem.”

For a quantum-scale system, however, it should be cautioned that even the expec­tation values do not truly satisfy Newtonian equations. Newtonian equations use the force at the expec­tation value of position, instead of the expec­tation value of the force. If the force varies non­linearly over the range of possible positions, it makes a difference.

There is a alternative formul­ation of quantum mechanics due to Heisenberg that is like the Ehrenfest theorem on steroids, {A.12}. Here the operators satisfy the Newtonian equations.

Key Points

$\begin{picture}(15,5.5)(0,-3) \put(2,0){\makebox(0,0){\scriptsize\bf0}} \put(12... ...\thicklines \put(3,0){\line(1,0){12}}\put(11.5,-2){\line(1,0){3}} \end{picture}$

Newtonian physics is an approximate version of quantum mechanics for macroscopic systems.

$\begin{picture}(15,5.5)(0,-3) \put(2,0){\makebox(0,0){\scriptsize\bf0}} \put(12... ...\thicklines \put(3,0){\line(1,0){12}}\put(11.5,-2){\line(1,0){3}} \end{picture}$

The equations of Newtonian physics apply to expec­tation values.

7.2.2 Energy-time uncertainty relation

The Heisenberg uncertainty relationship provides an intuitive way to understand the various weird features of quantum mechanics. The relationship says $\Delta{p}_x\Delta{x}$ $\raisebox{-.5pt}{$\geqslant$}$ $\frac12\hbar$, chapter 4.5.3. Here $\Delta{p}_x$ is the uncertainty in a component of the momentum of a particle, and $\Delta{x}$ is the uncertainty in the corre­sponding component of position.

Now special relativity considers the energy $E$ divided by the speed of light $c$ to be much like a zeroth momentum coordinate, and $ct$ to be much like a zeroth position coordinate, chapter 1.2.4 and 1.3.1. Making such substitutions transforms Heisenberg’s relationship into the so-called “energy-time uncertainty relationship:”

$$\fbox{$\displaystyle \Delta E\; \Delta t\mathrel{\raisebox{-1pt}{$\geqslant$}}\frac12\hbar $} %$$ (7.8)

There is a difference, however. In Heisenberg’s original relationship, the uncertainties in momentum and positions are mathemati­cally well defined. In particular, they are the standard deviations in the measurable values of these quantities. The uncertainty in energy in the energy-time uncertainty relationship can be defined similarly. The problem is what to make of that “uncertainty in time” $\Delta{t}$. The Schrödinger equation treats time fundamentally different from space.

One way to address the problem is to look at the typical evolution time of the expec­tation values of quantities of inter­est. Using careful analytical arguments along those lines, Mandelshtam and Tamm succeeded in giving a meaningful definition of the uncertainty in time, {A.18}. Unfortunately, its useful­ness is limited.

Ignore it. Careful analytical arguments are for wimps! Take out your pen and cross out “$\Delta{t}$.” Write in “any time difference you want.” Cross out “$\Delta{E}$” and write in “any energy difference you want.” As long as you are at it anyway, also cross out “$\raisebox{-.5pt}{$\geqslant$}$” and write in “$\vphantom0\raisebox{1.5pt}{$=$}$.” This can be justified because both are mathematical symbols. And inequalities are so vague anyway. You have now obtained the popular version of the Heisenberg energy-time uncertainty equality:

$$\fbox{$\displaystyle \mbox{any energy difference you wan... ...ifference you want} = {\textstyle\frac{1}{2}} \hbar $} %$$ (7.9)

This is an extremely powerful equation that can explain anything in quantum physics involving any two quantities that have dimensions of energy and time. Be sure, however, to only publicize the cases in which it gives the right answer.

Key Points

$\begin{picture}(15,5.5)(0,-3) \put(2,0){\makebox(0,0){\scriptsize\bf0}} \put(12... ...\thicklines \put(3,0){\line(1,0){12}}\put(11.5,-2){\line(1,0){3}} \end{picture}$

The energy-time uncertainty relationship is a gener­alization of the Heisenberg uncertainty relationship. It relates uncertainty in energy to uncertainty in time. What uncertainty in time means is not obvious.

$\begin{picture}(15,5.5)(0,-3) \put(2,0){\makebox(0,0){\scriptsize\bf0}} \put(12... ...\thicklines \put(3,0){\line(1,0){12}}\put(11.5,-2){\line(1,0){3}} \end{picture}$

If you are not a wimp, the answer to that problem is easy.