# Commutation Relations and Simultaneous Eigenvalues

## Commutator

The commutator of two operators $\hat{A}\!$ and $\hat{B}\!$ is defined as follows:

$[\hat{A},\hat{B}]=\hat{A}\hat{B}-\hat{B}\hat{A}\,\!.$

When 2 operators $\hat{A}\!$ and $\hat{B}\!$ commute, then $[\hat{A},\hat{B}]=0.$ On the other hand, if $[\hat{A},\hat{B}]\neq 0,$ then the operators do not commute, and we can think of the commutator between two operators as a measure of how badly they fail to commute. Note that any operator will commute with an ordinary complex number.

Some Identities:
$[\hat{A},\hat{B}]+[\hat{B},\hat{A}]=0 \!$
$[\hat{A},\hat{A}]= 0 \!$
$[\hat{A},\hat{B}+\hat{C}]=[\hat{A},\hat{B}]+[\hat{A},\hat{C}]\!$
$[\hat{A}+\hat{B},\hat{C}]=[\hat{A},\hat{C}]+[\hat{B},\hat{C}]\!$
$[\hat{A}\hat{B},\hat{C}]=\hat{A}[\hat{B},\hat{C}]+[\hat{A},\hat{C}]\hat{B}\!$
$[\hat{A},\hat{B}\hat{C}]=[\hat{A},\hat{B}]\hat{C}+\hat{B}[\hat{A},\hat{C}]\!$
$[\hat{A},[\hat{B},\hat{C}]]+[\hat{C},[\hat{A},\hat{B}]]+[\hat{B},[\hat{C},\hat{A}]]=0\!$
$[\hat{A},\hat{B}]=-[\hat{B},\hat{A}]\!$
$[\hat{A},\hat{B}]^{\dagger} = [\hat{B}^{\dagger},\hat{A}^{\dagger}]$

In addition, if any two operators are Hermitian and their product is also Hermitian, then the operators commute because

$(\hat{A}\hat{B})^{\dagger} = \hat{B}^{\dagger}\hat{A}^{\dagger} = \hat{B}\hat{A}$

and

$(\hat{A}\hat{B})^{\dagger} = \hat{A}\hat{B}$

so that $\hat{A}\hat{B} = \hat{B}\hat{A}.$

One may also prove that $[\hat{A}^n,\hat{B}]=n\hat{A}^{n-1}[\hat{A},\hat{B}]\!$ if $[\hat{A},[\hat{A},\hat{B}]]=0\!$ via mathematical induction.

## Compatible Observables

An operator which corresponds to some physically measurable property of a system is called an observable. The following is a list of common physical observables and their corresponding operators given in the coordinate representation:
Position: $\mathbf{r}\rightarrow\hat{\mathbf{r}}$
Momentum: $\mathbf{p}\rightarrow\frac{\hbar}{i}\nabla$
Kinetic energy: $T=\frac{p^2}{2m}\rightarrow -\frac{\hbar^2}{2m}\nabla^2$
Potential energy: $V(\mathbf{r})\rightarrow V(\hat{\mathbf{r}})$
Total energy: $E=T+V\rightarrow -\frac{\hbar^2}{2m}\nabla^2+V(\hat{\mathbf{r}})$

All observables are Hermitian.

Two observables $\hat{A}$ and $\hat{B}$ are said to be compatible if it is possible to exactly measure both simultaneously; i.e., they possess a common set of eigenstates:

$\hat{A}|\Psi_{AB}\rangle=a|\Psi_{AB}\rangle\!$
$\hat{B}|\Psi_{AB}\rangle=b|\Psi_{AB}\rangle\!$

It follows from this that

$\hat{A}\hat{B}|\Psi_{AB}\rangle=\hat{A}b|\Psi_{AB}\rangle=b\hat{A}|\Psi_{AB}\rangle=ba|\Psi_{AB}\rangle\!$

and

$\hat{B}\hat{A}|\Psi_{AB}\rangle=\hat{B}a|\Psi_{AB}\rangle=a\hat{B}|\Psi_{AB}\rangle=ba|\Psi_{AB}\rangle.\!$

Therefore,

$\hat{A}\hat{B}-\hat{B}\hat{A}=[\hat{A},\hat{B}]=0.\!$

The same logic works in reverse - if two operators commute, then they have simultaneous eigenkets, and thus they are compatible observables. Let us consider two commuting operators $\hat{A}$ and $\hat{B}.$ Suppose that we know the eigenstates of $\hat{A};$ let us call them $|\Psi_{A}\rangle.$ We know that

$\hat{A}|\Psi_{A}\rangle=a|\Psi_{A}\rangle.$

We now act on both sides of this equation with $\hat{B}:$

$\hat{B}\hat{A}|\Psi_{A}\rangle=a\hat{B}|\Psi_{A}\rangle,$

or, because $\hat{A}$ and $\hat{B}$ commute,

$\hat{A}\hat{B}|\Psi_{A}\rangle=a\hat{B}|\Psi_{A}\rangle.$

We see that $\hat{B}|\Psi_{A}\rangle$ is therefore also an eigenstate of $\hat{A}$ with the same eigenvalue. This means that we have "block diagonalized" $\hat{B}$ in the sense that it can only mix eigenstates of $\hat{A}$ with other eigenstates with the same eigenvalue. All of the eigenstates of $\hat{B}$ are thus also eigenstates of $\hat{A},$ meaning that the two observables are compatible.

These observations are especially important when we consider the Hamiltonian of a system. Symmetries of the Hamiltonian are represented by operators that commute with it, and, as we will see later, identification of these symmetries often helps one diagonalize the Hamiltonian and classify its eigenstates.

If $[\hat{A},\hat{B}]\neq 0$, or, equivalently, if it is not possible to simultaneously diagonalize $\hat{A}$ and $\hat{B},$ then the two operators are said to be incompatible observables.

## Generalized Uncertainty Relation

If two observables are incompatible, then one cannot necessarily simultaneously diagonalize them, so that an eigenstate of one may be a non-trivial linear combination of eigenstates of the other. This fact is embodied in the existence of an uncertainty relation for the two observables, much like that between position and momentum. We will now derive this general uncertainty relation.

For any observable A, let us define the uncertainty ΔA as the standard deviation of said observable from its expectation value:

$(\Delta A)^2 = \langle(A-\langle A\rangle)\Psi|(A-\langle A\rangle)\Psi\rangle = \langle f|f\rangle$

where $f \equiv (A-\langle A\rangle)\Psi$. Likewise, for any other observable B,

$(\Delta B)^2 = \langle g|g\rangle$ where $g \equiv (B-\langle B\rangle)\Psi$.

Now we invoke the Schwartz inequality. Recall that this is just

$\langle f|f\rangle\langle g|g\rangle\geq |\langle f|g\rangle |^2.$

Invoking this expression,

$(\Delta A)^2(\Delta B)^2 \geq |\langle f|g\rangle|^2.$

Now, for any complex number z,

$|z|^2 = [\text{Re}(z)]^2 + [\text{Im}(z)]^2 \geq [\text{Im}(z)]^2 = \left (\frac{z-z^*}{2i}\right )^2$.

Letting $z = \langle f|g\rangle$,

$(\Delta A)^2(\Delta B)^2 \geq \left (\frac{\langle f|g\rangle-\langle g|f\rangle}{2i}\right )^2$.

The inner products are

$\langle f|g\rangle = \langle(A-\langle A\rangle)\Psi|(B-\langle B\rangle)\Psi\rangle = \langle(A-\langle A\rangle)(B-\langle B\rangle)\rangle$
$= \langle AB - A\langle B\rangle - B\langle A\rangle + \langle A\rangle\langle B\rangle\rangle$
$= \langle AB\rangle - \langle B\rangle\langle A\rangle - \langle A\rangle\langle B\rangle + \langle A\rangle\langle B\rangle = \langle AB\rangle - \langle A\rangle\langle B\rangle$

and

$\langle g|f\rangle = \langle BA\rangle - \langle B\rangle\langle A\rangle.$

Therefore,

$\langle f|g\rangle - \langle g|f\rangle = \langle AB\rangle - \langle BA\rangle = \langle[A,B]\rangle$

and

$(\Delta A)^2(\Delta B)^2 \geq \left (\frac{1}{2i}\langle[A,B]\rangle\right )^2,$

or

$\Delta A\,\Delta B \geq \left |\frac{1}{2i}\langle[A,B]\rangle\right |.$

This is the general uncertainty relation that we sought.

As a simple demonstration of this relation, suppose that the first observable is position, A = x, and the second is momentum, $B=p=\frac{\hbar}{i}\frac{d}{dx}$.

The commutation relation between these two observables is just

$[x,p]=i\hbar,$

so

$\Delta x\,\Delta p\geq\frac{\hbar}{2}.$

We have thus obtained a formal proof of the Heisenberg Uncertainty Principle.

Question: What about the energy-time uncertainty relation?

Answer: We should note that time is not an operator in quantum mechanics, and thus we cannot apply the general uncertainty relation derived above to it. The energy-time uncertainty relation tells us that, for a short time interval, we get broadening in the energy spectrum which we observe. In other words, to get a precise energy value, we need to wait for a long time. The relation between the energy and time intervals is given by $\Delta E\,\Delta t\geq \hbar.$

## Functions of Operators

It is often useful to consider functions of operators, as the Hamiltonian can be written as such. For example, the exponential function of an operator is represented as a power series,

$e^\hat{A} = \hat{I} + \frac{\hat{A}}{1!} + \frac{\hat{A}^2}{2!}+ \frac{\hat{A}^3}{3!} + \ldots$

This leads to the identity that for some $f(\lambda) = e^{\lambda\hat{A}}\hat{B}e^{-\lambda\hat{A}}$, where $\hat{A}$ and $\hat{B}$ are operators:

$e^{\lambda\hat{A}}\hat{B}e^{-\lambda\hat{A}} = \hat{B} + \frac{\lambda}{1!}[\hat{A},\hat{B}] + \frac{\lambda^2}{2!}[\hat{A},[\hat{A},\hat{B}]] + \frac{\lambda^3}{3!}[\hat{A},[\hat{A},[\hat{A},\hat{B}]]] + \ldots$

This follows easily when one considers a Taylor series expansion of $f(\lambda).\!$ We note that

$\frac{d}{d\lambda}f(\lambda)=e^{\lambda\hat{A}}[\hat{A},\hat{B}]e^{-\lambda\hat{A}}.$

We may then apply this relation recursively to generate the higher-order derivatives required in the Taylor expansion.

If $[\hat{A},[\hat{A},\hat{B}]]=\beta\hat{B},$ then we can simplify the identity to

$e^{\lambda\hat{A}}\hat{B}e^{-\lambda\hat{A}}=\hat{B}\cosh\lambda\sqrt{\beta}+\frac{[\hat{A},\hat{B}]}{\sqrt{\beta}}\sinh\lambda\sqrt{\beta}.$

It is also useful to consider the commutator of an operator function with an operator. Given operators $\hat{l}$ and $\hat{m}$ where

$[\hat{l},\hat{m}] = 1$

it can be shown that, for arbitrary $f(\hat{l})$,

$[f(\hat{l}),\hat{m}] = \frac{d}{d\hat{l}}f(\hat{l})$

by use of the power series expansion of $f(\hat{l})$ and the commutator identities in the above section.

## Physical Meaning of Various Representations

In a given Hilbert basis, it is obvious that the state vector $|\psi \rangle$ is completely determined by the set of its components $\{c_n\},\!$

$|\psi\rangle\leftrightarrow\left \{c_{n}=\langle\phi_{n}|\psi\rangle\right \}$

which we can write as a column vector; the corresponding bra is then represented by the conjugate line vector. This representation of the state vector is completely equivalent to the wave function $\Psi(\mathbf{r},t).$

Therefore, there are not only two, but an infinite number of equivalent representations of the state of the system. What is their physical meaning?

In the basis of the eigenstates of the Hamiltonian, the interpretation of this representation is simple and crystal clear: the coefficients $c_{n}\!$ are the probability amplitudes for obtaining $E_{n}\!$ in an energy measurement.

Therefore,

• The position representation, $\Psi(\mathbf{r},t),$ is more convenient if we are interested in the properties of the particle in space.

• Its Fourier transform, $\Phi(\mathbf{p},t),$ is more convenient if we are interested in its momentum properties.

• The coefficients $c_n\!$ in the basis of energy eigenstates are more convenient if we are interested in the energy of the particle.

Owing to Riesz’s theorem, this can be done with any observable, such as the angular momentum, which we examine later on and also has discrete eigenvalues. This can be thought of as a “generalization” of the Fourier transform.

## Position and Momentum Operators

An extremely useful example is the commutation relation of the position $\hat{x}$ and momentum $\hat{p}$ operators. In the position representation, the position operator is $\hat{x}\rightarrow x$ and the momentum operator is $\hat{p}\rightarrow \frac{\hbar}{i}\frac{\partial}{\partial x}$. Similarly, in the momentum representation, the momentum operator is $\hat{p}\rightarrow p$ and position operator $\hat{x}\rightarrow i\hbar\frac{\partial}{\partial p}$.

Applying $\hat{x}$ and $\hat{p}$ to an arbitrary state vector, we can see that

$\left [\hat{x},\hat{p}\right ]= i\hbar.$

The position and momentum operators are thus incompatible. This provides a fundamental contrast to classical mechanics in which x and p obviously commute. Such a commutation relation holds for any coordinate and its canonically conjugate momentum; the commutation relations for all coordinates and their corresponding momenta are known as the canonical commutation relations.

In three dimensions, the canonical commutation relations are

$\left[\hat{x}_i,\hat{p}_j\right]= i\hbar\delta_{ij},$

$\left[\hat{x}_i,\hat{x}_j\right]=0,$

and

$\left[\hat{p}_i,\hat{p}_j\right]=0,$

where the indices stand for the x, y, and z components of the 3-vectors.

It is again interesting to consider functions of these operators. For some function of a coordinate $\hat{q},$ $f(\hat{q}),$ it is easily shown that

$[\hat{p},f(\hat{q})] = -i\hbar\frac{\partial}{\partial q}f(\hat{q}).$

Additionally, the expression $e^{i\lambda\hat{p}/\hbar}f(\hat{q})e^{-i\lambda\hat{p}/\hbar}=f(\hat{q}+\lambda);$ i.e., the operator $\hat{T}(\lambda)=e^{i\lambda\hat{p}/\hbar}$ is a translation operator that shifts the coordinate $\hat{q}$ by an amount $\lambda.\!$

## Connection between Classical and Quantum Mechanics

There is a wonderful connection between classical and quantum mechanics. The Hamiltonian is a function that appears in classical mechanics and is "promoted" to an operator in quantum mechanics. Classically, the Hamiltonian is defined as

$H(q,p;t)=\sum_kp_{k}\dot{q}_{k}-L(q,\dot{q};t),$

where $L(q,\dot{q};t)$ is the Lagrangian. There are two properties of this quantity.

1. If L does not depend explicitly on time, then H is conserved.
2. Furthermore, if the potential energy and the constraints of the system are time-independent, then H is not only conserved, but H is the energy of the system.

The equations of motion for the system that we may derive from the Hamiltonian are

$\dot{q}_k=\frac{\partial H}{\partial p_{k}}$

and

$\dot{p}_k=-\frac{\partial H}{\partial q_{k}}.$

These equations are called Hamilton's equations of motion. We also define the quantity,

$[A,B]=\sum_{k}\left (\frac{\partial A}{\partial q_{k}}\frac{\partial B}{\partial p_{k}}-\frac{\partial A}{\partial p_{k}}\frac{\partial B}{\partial q_{k}}\right ),$

known as the Poisson bracket. We may observe an interesting connection between this purely classical concept and the commutators that we encounter in quantum mechanics as follows. Let us calculate this quantity with A and B different coordinates and momenta.

$[p_i,p_j]=\sum_{k}\left(\frac{\partial p_i}{\partial q_{k}}\frac{\partial p_j}{\partial p_{k}}- \frac{\partial p_i}{\partial p_{k}}\frac{\partial p_j}{\partial q_{k}}\right)=0$

$[q_i,q_j]=\sum_{k}\left(\frac{\partial q_i}{\partial q_{k}}\frac{\partial q_j}{\partial p_{k}}- \frac{\partial q_i}{\partial p_{k}}\frac{\partial q_j}{\partial q_{k}}\right)=0$

$[p_i,q_j]=\sum_{k}\left(\frac{\partial p_i}{\partial q_{k}}\frac{\partial q_j}{\partial p_{k}}- \frac{\partial p_i}{\partial p_{k}}\frac{\partial q_j}{\partial q_{k}}\right)=\sum_{k}-\delta_{ik}\delta_{jk}=-\delta_{ij}.$

These relations are very similar to the canonical commutation relations discussed earlier. In fact, if we make the replacement,

$\delta_{ij}\rightarrow i\hbar\delta_{ij},$

and replace the Poisson brackets with commutators, then we obtain the canonical commutation relations. This identification is called canonical quantization.

As a final important remark, there is no classical analogue to the Heisenberg uncertainty relation for conjugate classical variables in terms of, say, their Poisson brackets. This is because, in classical mechanics, the objects that we study are point particles (or collections thereof) with well-defined positions and momenta. In quantum mechanics, on the other hand, we study the state of a system as encoded in its wave function. This wave function describes the probability of finding a particle at a given position or with a given momentum; the particles that make up the system do not have definite positions or momenta.

## Problems

(1) Let $f(x) \!$ be a differentiable function. Using the fact that $[\hat{x},\hat{p}_{x}]=i\hbar,$ prove the following identities:

(a) $[\hat{x},\hat{p}^{2}_{x}f(\hat{x}) ]=2i\hbar \hat{p}_{x}f(\hat{x})$

(b) $[\hat{x},\hat{p}_{x}f(\hat{x})\hat{p}_{x}]=i\hbar[f(\hat{x})\hat{p}_{x}+\hat{p}_{x}f(\hat{x})]$

(c) $[\hat{p}_{x},\hat{p}^{2}_{x}f(\hat{x})]=-i\hbar\hat{p}^{2}_{x}\frac{df(\hat{x})}{dx}$

(d) $[\hat{p}_{x},\hat{p}_{x}f(\hat{x})\hat{p}_{x}]=-i\hbar\hat{p}_{x}\frac{df(\hat{x})}{dx}\hat{p}_{x}$

(2) (From Sakurai, Modern Quantum Mechanics, Problem 1.30)

The translation operator for a finite (spatial) displacement $\mathbf{l}$ is given by $\hat{T}(\mathbf{l})=\exp\left (-\frac{i\hat{\mathbf{p}}\cdot\mathbf{l}}{\hbar}\right ),$ where $\hat{\mathbf{p}}$ is the momentum operator.

(a) Evaluate $[\hat{x}_{i},\hat{T}(\mathbf{l})].$

(b) Using your result from (a), demonstrate how the expectation value $\langle\hat{\mathbf{x}}\rangle$ changes under translation.