Commutators

Definition¶

Let $A, B$ be two operators. The commutator of $A$ and $B$ is given by

This means multiply the operator $A$ on the left by $B$, and then taking off the operator $B$ multiplied on the left by $A$.

When considering the matrix decomposition of these operators, $A'$ and $B'$, this relation can be seen as matrices for which their ordered multiplication is not equal: $A'B' \neq B'A'$.

Commutator Identities¶

Here are some common commutator identities that are useful.

Let $A,B,C$ be some operators and $\mathbb{I}$ be the identity operator.

1. $[A,A] = 0$.
2. $[A, \mathbb{I}] = 0$.
3. $[A,B]=-[B,A]$.
4. $[A,BC] = [A,B]C + B[A,C]$.
5. $[AB,C] = A[B,C] + [A,C]B$.

Mathematical Implications¶

When two observables $A$ and $B$ commute, $[A,B] = 0$, they share a common eigenbasis.

Non-Degenerate Operators¶

If $A$ and $B$ are non-degenerate operators then the shared eigenbasis is unique.

That is, each non-degenerate eigenvector of $A$ is also an eigenvector of $B$.

degenerate Operators¶

If $A$ is degenerate then in the degenerate sub-space there are an infinite number of vectors that are eigenvectors of $A$ - namely, any linear combination of eigenvectors in the degenerate eigenspace is also an eigenvector $A$. A common eigenbasis can still be found but it is in general not unique.

For each degenerate eigenvalue $\alpha$ of $A$, of multiplicity $n$, one can find a set of $n$ vectors, $\{ \ket{\phi_m} \}_{m=0}^{n-1}$, such that

Hence, each $\ket{\phi_m}$ is an eigenvector of both $A$ and $B$, but not all eigenvectors of $A$ corresponding to the degenerate eigenvalue $\alpha$ are eigenvectors of $B$.

Nothing can be said about the eigenvalues, $\lambda_m$, from this theory.

Physical Implication¶

The fact that there exists observables for which $[A,B] \neq 0$ is the origin of much that is interesting about quantum mechanics. Physically, it means that measuring $A$ and then $B$ on a state $\ket{\psi}$ will give a different outcome to measuring $B$ and then $A$ on $\ket{\psi}$. This is due to measurement physically changing the underlying state.

The consequence of this is that it is not possible to simultaneously measure the observables $A$ and $B$ if they do not commute. For example, there exists sets of states and measurements for which a measurement of $A$ means one is completely uncertain about what measurement outcome they would get if they were to then measure $B$. In this case, measuring $A$ (and hence knowing it with certainty) makes one maximally uncertain about what measurement outcome they would get when measuring $B$.

To see why a shared eigenbasis allows simultaneous measurement, consider a state, $\ket{\psi}$, upon which the observable $A$ is measured where $[A,B] = 0$. If $A=\sum j \ket{j}\bra{j}$, then the measurement outcome will be some eigenvalue $k$ and the post measurement state will be the associated eigenvector $\ket{k}$. Considering now that the observable $B$ is then measured on the same system. As $A$ and $B$ share a common eigenbasis, $\ket{k}$ is also an eigenvector of $B$ and hence the measurement outcome will be $b_k$, where $B\ket{k}=b_k\ket{k}$, with certainty. Therefore, if one knows the measurement outcome from measuring $A$, they know for certain what the measurement outcome from measuring $B$ will be. Moreover, if they were to measure $A$ again, they would again get $k$ with certainty. Hence, they can know both measurement outcomes simultaneously with certainty.

If the observables do not commute, $[A,B] \neq 0$, they do not share a common eigenbasis. After measuring $A$, the state will be in an eigenvector of the observable $A$. After measuring $B$, the state will be in an eigenvector of the observable $B$. Each measurement projects the state onto a different basis, with the output state being probabilistic each time. cycling through measurements of $A, B, A, B \ldots$ could therefore give different measurement outcomes each time the same observable is measured.