# Time evolution of observables

### Evolution of averages

An observable of some given system doesn't, in general, change with time. That is, if $A$ is Hermitian then $$\frac{\partial A}{\partial t} \equiv 0$$

But, the states that the system can be in do change with time, as shown already, and the change is related to the Hamiltonian of the system, $$\frac{\partial}{\partial t} \ket{\psi} = -\mathrm{i}\frac{H}{\hbar} \ket{\psi}$$

The conjugate version of this result is $$\frac{\partial}{\partial t} \bra{\psi} = \bra{\psi}\frac{H}{\hbar}\mathrm{i}$$

Thus, we can calculate the time rate of change of the average of $A$ in the state $\ket{\psi}$, \begin{align*}\frac{\partial}{\partial t} \left\langle A \right\rangle&= \frac{\partial}{\partial t} \bra{\psi}A\ket{\psi}\\&= \left(\frac{\partial}{\partial t}\bra{\psi}\right) A\ket{\psi} + \bra{\psi}A \left(\frac{\partial}{\partial t}\ket{\psi}\right)\\&= \left(\bra{\psi}\frac{H}{\hbar}\mathrm{i}\right) A\ket{\psi} + \bra{\psi}A \left(-\mathrm{i}\frac{H}{\hbar} \ket{\psi}\right)\\&= \frac{\mathrm{i}}{\hbar} \bra{\psi}HA - AH\ket{\psi}\\&= \frac{\mathrm{i}}{\hbar} \left\langle HA - AH \right\rangle\end{align*}

The term $$\left[H,A\right] = HA - AH$$ is called the commutator of $H$ with $A$.

Nb. Commutators crop up all over physics. In terms of Hermitian operators, it is easy to show that, for example that, for any $A$, $\left[A,A\right] = 0$. More generally, $\left[A,B\right] = 0$ implies that $A$ and $B$ commute, and so can be measured simultaneously.

So we can write the derivative in terms of a commutator, $$\frac{\partial}{\partial t} \left\langle A \right\rangle = \frac{\mathrm{i}}{\hbar} \left\langle \left[H,A\right] \right\rangle$$

In English, the derivative with respect to time of the average of an observable is proportional to the average of the commutator of the Hamiltonian with the observable, and the constant of proportionality is $\frac{\mathrm{i}}{\hbar}$.

Professor Susskind makes the point that we usually measuring the averages of observables - rather than the observables themselves - so that, when the context is clear, we can drop the brackets from the above result and write $$\frac{\partial A}{\partial t} = \frac{\mathrm{i}}{\hbar}\left[H,A\right]$$

### Conservation laws

First we note that average of the Hamiltonian, $H$, of a system is a conserved quantity.

To see why this is true, simply set $A = H$ in the equation for the derivative of the average, which we can do because the Hamiltonian is an observable. Then, $$\frac{\partial}{\partial t}\left\langle H \right\rangle = \frac{\mathrm{i}}{\hbar}\left\langle \left[H,H\right] \right\rangle = \frac{\mathrm{i}}{\hbar}\left\langle \left(HH - HH\right) \right\rangle = 0$$

The eigenvalues of the Hamiltonian are, in most cases, the discrete energy bands that a system can attain. So the law of conservation of the Hamiltonian is a generalisation of the law of conservation of energy.

Further, as noted above, if a pair of observables commute, then their commutator is zero. Hence, for any operator $A$, that commutes with the Hamiltonian $H$, $$\frac{\partial}{\partial t}\left\langle A \right\rangle = \frac{\mathrm{i}}{\hbar}\left\langle \left[H, A\right] \right\rangle = \frac{\mathrm{i}}{\hbar}\left\langle 0 \right\rangle = 0$$

Thus we have another conservation law - the conservation of averages of observables that commute with the Hamiltonian.

In particular, any observable that has the same set of eigenvectors as the Hamiltonian is a conserved quantity.

Nb. This is because, as matrices, both the observable and the Hamiltonian will be diagonal on the basis of common eigenvectors, and diagonal matrices commute.