Subsections


4.5 The Commutator

As the previous section discussed, the standard deviation $\sigma$ is a measure of the uncertainty of a property of a quantum system. The larger the standard deviation, the farther typical measurements stray from the expected average value. Quantum mechanics often requires a minimum amount of uncertainty when more than one quantity is involved, like position and linear momentum in Heisenberg's uncertainty principle. In general, this amount of uncertainty is related to an important mathematical object called the commutator, to be discussed in this section.


4.5.1 Commuting operators

First, note that there is no fundamental reason why several quantities cannot have a definite value at the same time. For example, if the electron of the hydrogen atom is in a $\psi_{nlm}$ eigenstate, its total energy, square angular momentum, and $z$-​component of angular momentum all have definite values, with zero uncertainty.

More generally, two different quantities with operators $A$ and $B$ have definite values if the wave function is an eigenfunction of both $A$ and $B$. So, the question whether two quantities can be definite at the same time is really whether their operators $A$ and $B$ have common eigenfunctions. And it turns out that the answer has to do with whether these operators “commute”, in other words, on whether their order can be reversed as in $AB$ $\vphantom0\raisebox{1.5pt}{$=$}$ $BA$.

In particular, {D.18}:

Iff two Hermitian operators commute, there is a complete set of eigenfunctions that is common to them both.
(For more than two operators, each operator has to commute with all others.)

For example, the operators $H_x$ and $H_y$ of the harmonic oscillator of chapter 4.1.2 commute:

\begin{eqnarray*}
H_x H_y \Psi & = &
\left[
- \frac{\hbar^2}{2m} \frac{\pa...
...x^2 {\textstyle\frac{1}{2}} cy^2\Psi \\
& = &
H_y H_x \Psi
\end{eqnarray*}

This is true since it makes no difference whether you differentiate $\Psi$ first with respect to $x$ and then with respect to $y$ or vice versa, and since the $\frac12cy^2$ can be pulled in front of the $x$-​differentiations and the $\frac12cx^2$ can be pushed inside the $y$-​differentiations, and since multiplications can always be done in any order.

The same way, $H_z$ commutes with $H_x$ and $H_y$, and that means that $H$ commutes with them all, since $H$ is just their sum. So, these four operators should have a common set of eigenfunctions, and they do: it is the set of eigenfunctions $\psi_{n_xn_yn_z}$ derived in chapter 4.1.2.

Similarly, for the hydrogen atom, the total energy Hamiltonian $H$, the square angular momentum operator $\L ^2$ and the $z$-​component of angular momentum $\L _z$ all commute, and they have the common set of eigenfunctions $\psi_{nlm}$.

Note that such eigenfunctions are not necessarily the only game in town. As a counter-example, for the hydrogen atom $H$, $\L ^2$, and the $x$-​component of angular momentum $\L _x$ also all commute, and they too have a common set of eigenfunctions. But that will not be the $\psi_{nlm}$, since $\L _x$ and $\L _z$ do not commute. (It will however be the $\psi_{nlm}$ after you rotate them all 90 degrees around the $y$-​axis.) It would certainly be simpler mathematically if each operator had just one unique set of eigenfunctions, but nature does not cooperate.


Key Points
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Operators commute if you can change their order, as in $AB$ $\vphantom0\raisebox{1.5pt}{$=$}$ $BA$.

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For commuting operators, a common set of eigenfunctions exists.

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For those eigenfunctions, the physical quantities corresponding to the commuting operators all have definite values at the same time.

4.5.1 Review Questions
1.
The pointer state

\begin{displaymath}
\mbox{2p$_x$} = \frac 1{\sqrt 2}\left(-\psi_{211}+\psi_{21-1}\right).
\end{displaymath}

is one of the eigenstates that $H$, $\L ^2$, and $\L _x$ have in common. Check that it is not an eigenstate that $H$, $\L ^2$, and $\L _z$ have in common.

Solution commutea-a


4.5.2 Noncommuting operators and their commutator

Two quantities with operators that do not commute cannot in general have definite values at the same time. If one has a definite value, the other is in general uncertain.

The qualification in general is needed because there may be exceptions. The angular momentum operators do not commute, but it is still possible for the angular momentum to be zero in all three directions. But as soon as the angular momentum in any direction is nonzero, only one component of angular momentum can have a definite value.

A measure for the amount to which two operators $A$ and $B$ do not commute is the difference between $AB$ and $BA$; this difference is called their “commutator” $[A,B]$:

\begin{displaymath}
\fbox{$\displaystyle [A,B] \equiv A B - B A $}
\end{displaymath} (4.45)

A nonzero commutator $[A,B]$ demands a minimum amount of uncertainty in the corresponding quantities $a$ and $b$. It can be shown, {D.19}, that the uncertainties, or standard deviations, $\sigma_a$ in $a$ and $\sigma_b$ in $b$ are at least so large that:

\begin{displaymath}
\fbox{$\displaystyle
\sigma_a \sigma_b \mathrel{\raisebo...
...}{\textstyle\frac{1}{2}} \vert\langle[A,B]\rangle\vert
$} %
\end{displaymath} (4.46)

This equation is called the “generalized uncertainty relationship”.


Key Points
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The commutator of two operators $A$ and $B$ equals $AB-BA$ and is written as $[A,B]$.

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The product of the uncertainties in two quantities is at least one half the magnitude of the expectation value of their commutator.


4.5.3 The Heisenberg uncertainty relationship

This section will work out the uncertainty relationship (4.46) of the previous subsection for the position and linear momentum in an arbitrary direction. The result will be a precise mathematical statement of the Heisenberg uncertainty principle.

To be specific, the arbitrary direction will be taken as the $x$-​axis, so the position operator will be ${\widehat x}$, and the linear momentum operator ${\widehat p}_x$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\hbar\partial$$\raisebox{.5pt}{$/$}$${\rm i}\partial{x}$. These two operators do not commute, ${\widehat p}_x{\widehat x}\Psi$ is simply not the same as ${\widehat x}{\widehat p}_x\Psi$: ${\widehat p}_x{\widehat x}\Psi$ means multiply function $\Psi$ by $x$ to get the product function $x\Psi$ and then apply ${\widehat p}_x$ on that product, while ${\widehat x}{\widehat p}_x\Psi$ means apply ${\widehat p}_x$ on $\Psi$ and then multiply the resulting function by $x$. The difference is found from writing it out:

\begin{displaymath}
{\widehat p}_x {\widehat x}\Psi
= \frac{\hbar}{{\rm i}} ...
...l x}
= -{\rm i}\hbar \Psi + {\widehat x}{\widehat p}_x \Psi
\end{displaymath}

the second equality resulting from differentiating out the product.

Comparing start and end shows that the difference between ${\widehat x}{\widehat p}_x$ and ${\widehat p}_x{\widehat x}$ is not zero, but ${\rm i}\hbar$. By definition, this difference is their commutator:

\begin{displaymath}
\fbox{$\displaystyle [{\widehat x},{\widehat p}_x] = {\rm i}\hbar $} %
\end{displaymath} (4.47)

This important result is called the “canonical commutation relation.” The commutator of position and linear momentum in the same direction is the nonzero constant ${\rm i}\hbar$.

Because the commutator is nonzero, there must be nonzero uncertainty involved. Indeed, the generalized uncertainty relationship of the previous subsection becomes in this case:

\begin{displaymath}
\sigma_{x} \sigma_{p_x} \mathrel{\raisebox{-1pt}{$\geqslant$}}{\textstyle\frac{1}{2}} \hbar
\end{displaymath} (4.48)

This is the uncertainty relationship as first formulated by Heisenberg.

It implies that when the uncertainty in position $\sigma_{x}$ is narrowed down to zero, the uncertainty in momentum $\sigma_{p_x}$ must become infinite to keep their product nonzero, and vice versa. More generally, you can narrow down the position of a particle and you can narrow down its momentum. But you can never reduce the product of the uncertainties $\sigma_{x}$ and $\sigma_{p_x}$ below $\frac12\hbar$, whatever you do.

It should be noted that the uncertainty relationship is often written as $\Delta{p}_x\Delta{x}$ $\raisebox{-.5pt}{$\geqslant$}$ $\frac12\hbar$ or even as $\Delta{p_x}\Delta{x}$ $\vphantom0\raisebox{1.1pt}{$\approx$}$ $\hbar$ where $\Delta{p}$ and $\Delta{x}$ are taken to be vaguely described uncertainties in momentum and position, rather than rigorously defined standard deviations. And people write a corresponding uncertainty relationship for time, $\Delta{E}\Delta{t}$ $\raisebox{-.5pt}{$\geqslant$}$ $\frac12\hbar$, because relativity suggests that time should be treated just like space. But note that unlike the linear momentum operator, the Hamiltonian is not at all universal. So, you might guess that the definition of the uncertainty $\Delta{t}$ in time would not be universal either, and you would be right, chapter 7.2.2.


Key Points
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The canonical commutator $[{\widehat x},{\widehat p}_x]$ equals ${\rm i}\hbar$.

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If either the uncertainty in position in a given direction or the uncertainty in linear momentum in that direction is narrowed down to zero, the other uncertainty blows up.

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The product of the two uncertainties is at least the constant $\frac12\hbar$.

4.5.3 Review Questions
1.
This sounds serious! If I am driving my car, the police requires me to know my speed (linear momentum). Also, I would like to know where I am. But neither is possible according to quantum mechanics.

Solution commutec-a


4.5.4 Commutator reference

It is a fact of life in quantum mechanics that commutators pop up all over the place. Not just in uncertainty relations, but also in the time evolution of expectation values, in angular momentum, and in quantum field theory, the advanced theory of quantum mechanics used in solids and relativistic applications. This section can make your life easier dealing with them. Browse through it to see what is there. Then come back when you need it.

Recall the definition of the commutator $[A,B]$ of any two operators $A$ and $B$:

\begin{displaymath}[A,B]= AB - BA %
\end{displaymath} (4.49)

By this very definition , the commutator is zero for any two operators $A_1$ and $A_2$ that commute, (whose order can be interchanged):
\begin{displaymath}[A_1,A_2]= 0 \quad
\mbox{if $A_1$ and $A_2$ commute; } A_1 A_2 = A_2 A_1. %
\end{displaymath} (4.50)

If operators all commute, all their products commute too:
\begin{displaymath}[A_1 A_2 \ldots A_k, A_{k+1}\ldots A_n]= 0
\quad \mbox{if }...
..._2, \ldots, A_k, A_{k+1},\ldots, A_n
\mbox{ all commute.} %
\end{displaymath} (4.51)

Everything commutes with itself, of course:

\begin{displaymath}[A,A]= 0, %
\end{displaymath} (4.52)

and everything commutes with a numerical constant; if $A$ is an operator and $a$ is some number, then:
\begin{displaymath}[A,a]= [a,A] = 0. %
\end{displaymath} (4.53)

The commutator is antisymmetric; or in simpler words, if you interchange the sides; it will change the sign, {D.20}:

\begin{displaymath}[B,A]= - [A,B]. %
\end{displaymath} (4.54)

For the rest however, linear combinations multiply out just like you would expect:
\begin{displaymath}[aA+bB,cC+dD]=
ac[A,C] + ad[A,D] + bc[B,C] + bd[B,D], %
\end{displaymath} (4.55)

(in which it is assumed that $A$, $B$, $C$, and $D$ are operators, and $a$, $b$, $c$, and $d$ numerical constants.)

To deal with commutators that involve products of operators, the rule to remember is: “the first factor comes out at the front of the commutator, the second at the back”. More precisely:

\begin{displaymath}
\rule[-10pt]{7pt}{0pt}
\raisebox{-7pt}{
\begin{picture...
...end{picture}
}
[\ldots,AB] = A[\ldots,B] + [\ldots,A]B. %
\end{displaymath} (4.56)

So, if $A$ or $B$ commutes with the other side of the operator, it can simply be taken out at at its side; (the second commutator will be zero.) For example,

\begin{displaymath}[A_1B, A_2]= A_1 [B, A_2], \qquad [B A_1, A_2] = [B, A_2] A_1
\end{displaymath}

if $A_1$ and $A_2$ commute.

Now from the general to the specific. Because changing sides in a commutator merely changes its sign, from here on only one of the two possibilities will be shown. First the position operators all mutually commute:

\begin{displaymath}[{\widehat x},{\widehat y}]=[{\widehat y},{\widehat z}]=[{\widehat z},{\widehat x}]=0 %
\end{displaymath} (4.57)

as do position-dependent operators such as a potential energy $V(x,y,z)$:
\begin{displaymath}[{\widehat x},V(x,y,z)]=[{\widehat y},V(x,y,z)]=[{\widehat z},V(x,y,z)]=0 %
\end{displaymath} (4.58)

This illustrates that if a set of operators all commute, then all combinations of those operators commute too.

The linear momentum operators all mutually commute:

\begin{displaymath}[{\widehat p}_x,{\widehat p}_y]=
[{\widehat p}_y,{\widehat p}_z] =
[{\widehat p}_z,{\widehat p}_x] = 0 %
\end{displaymath} (4.59)

However, position operators and linear momentum operators in the same direction do not commute; instead:
\begin{displaymath}[{\widehat x},{\widehat p}_x]=
[{\widehat y},{\widehat p}_y] =
[{\widehat z},{\widehat p}_z] = {\rm i}\hbar %
\end{displaymath} (4.60)

As seen in the previous subsection, this lack of commutation causes the Heisenberg uncertainty principle. Position and linear momentum operators in different directions do commute:
\begin{displaymath}[{\widehat x},{\widehat p}_y]=
[{\widehat x},{\widehat p}_z...
...at z},{\widehat p}_x] =
[{\widehat z},{\widehat p}_y] = 0 %
\end{displaymath} (4.61)

A generalization that is frequently very helpful is:

\begin{displaymath}[f,{\widehat p}_x]= {\rm i}\hbar \frac{\partial f}{\partial x...
...\widehat p}_z] = {\rm i}\hbar \frac{\partial f}{\partial z} %
\end{displaymath} (4.62)

where $f$ is any function of $x$, $y$, and $z$.

Unlike linear momentum operators, angular momentum operators do not mutually commute. The commutators are given by the so-called “ fundamental commutation relations:”

\begin{displaymath}[\L _x,\L _y]= {\rm i}\hbar\L _z \quad
[\L _y,\L _z] = {\rm i}\hbar\L _x \quad
[\L _z,\L _x] = {\rm i}\hbar\L _y %
\end{displaymath} (4.63)

Note the $\ldots{x}yzxyz\ldots$ order of the indices that produces positive signs; a reversed $\ldots{z}yxzy\ldots$ order adds a minus sign. For example $[\L _z,\L _y]$ $\vphantom0\raisebox{1.5pt}{$=$}$ $-{\rm i}\hbar\L _x$ because $y$ following $z$ is in reversed order.

The angular momentum components do all commute with the square angular momentum operator:

\begin{displaymath}[\L _x,\L ^2]= [\L _y,\L ^2] = [\L _z,\L ^2] = 0
\quad\mbox{where } \L ^2 = \L _x^2 + \L _y^2 + \L _z^2 %
\end{displaymath} (4.64)

Just the opposite of the situation for linear momentum, position and angular momentum operators in the same direction commute,

\begin{displaymath}[{\widehat x}, \L _x]= [{\widehat y}, \L _y] = [{\widehat z}, \L _z] = 0 %
\end{displaymath} (4.65)

but those in different directions do not:
\begin{displaymath}[{\widehat x},\L _y]=[\L _x,{\widehat y}]={\rm i}\hbar {\wide...
...at z},\L _x]=[\L _z,{\widehat x}]={\rm i}\hbar {\widehat y} %
\end{displaymath} (4.66)

Square position commutes with all components of angular momentum,
\begin{displaymath}[{\widehat r}^2,\L _x]= [{\widehat r}^2,\L _y] = [{\widehat r}^2,\L _z]
= [{\widehat r}^2,\L ^2] = 0 %
\end{displaymath} (4.67)

The commutator between position and square angular momentum is, using vector notation for conciseness,
\begin{displaymath}[{\skew 2\widehat{\skew{-1}\vec r}},\L ^2]= - 2\hbar^2 {\skew...
...{\skew{-1}\vec r}}\cdot{\skew 4\widehat{\skew{-.5}\vec p}}) %
\end{displaymath} (4.68)

The commutators between linear and angular momentum are very similar to the ones between position and angular momentum:

\begin{displaymath}[{\widehat p}_x, \L _x]= [{\widehat p}_y, \L _y] = [{\widehat p}_z, \L _z] = 0 %
\end{displaymath} (4.69)


\begin{displaymath}[{\widehat p}_x,\L _y]=[\L _x,{\widehat p}_y]={\rm i}\hbar {\...
...z,\L _x]=[\L _z,{\widehat p}_x]={\rm i}\hbar {\widehat p}_y %
\end{displaymath} (4.70)


\begin{displaymath}[{\widehat p}^2,\L _x]= [{\widehat p}^2,\L _y] = [{\widehat p}^2,\L _z]
= [{\widehat p}^2,\L ^2] = 0 %
\end{displaymath} (4.71)


\begin{displaymath}[{\skew 4\widehat{\skew{-.5}\vec p}},\L ^2]= - 2\hbar^2 {\ske...
...\skew{-.5}\vec p}}\cdot{\skew 4\widehat{\skew{-.5}\vec p}}) %
\end{displaymath} (4.72)

The following commutators are also useful:

\begin{displaymath}[{\skew0\vec r}\times{\skew 4\widehat{\vec L}},\L ^2]= 2 {\rm...
...,\L ^2] = 2\hbar^2({\skew0\vec r}\L ^2+\L ^2{\skew0\vec r}) %
\end{displaymath} (4.73)

Commutators involving spin are discussed in a later chapter, 5.5.3.


Key Points
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Rules for evaluating commutators were given.

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Return to this subsection if you need to figure out some commutator or the other.