# College Math Teaching

## August 8, 2011

### Quantum Mechanics and Undergraduate Mathematics VIII: Time Evolution of Expectation of an Observable

Filed under: advanced mathematics, applied mathematics, physics, probability, quantum mechanics, science — collegemathteaching @ 3:12 pm

Back to our series on QM: one thing to remember about observables: they are operators with a set collection of eigenvectors and eigenvalues (allowable values that can be observed; “quantum levels” if you will). These do not change with time. So $\frac{d}{dt} (A (\psi)) = A (\frac{\partial}{\partial t} \psi)$. One can work this out by expanding $A \psi$ if one wants to.

So with this fact, lets see how the expectation of an observable evolves with time (given a certain initial state):
$\frac{d}{dt} E(A) = \frac{d}{dt} \langle \psi, A \psi \rangle = \langle \frac{\partial}{\partial t} \psi, A \psi \rangle + \langle \psi, A \frac{\partial}{\partial t} \psi \rangle$

Now apply the Hamiltonian to account for the time change of the state vector; we obtain:
$\langle -\frac{i}{\hbar}H \psi, A \psi \rangle + \langle \psi, -\frac{i}{\hbar}AH \psi \rangle = \overline{\frac{i}{\hbar}} \langle H \psi, A \psi \rangle + -\frac{i}{\hbar} \langle \psi, AH \psi \rangle$

Now use the fact that both $H$ and $A$ are Hermitian to obtain:
$\frac{d}{dt} A = \frac{i}{\hbar} \langle \psi, (HA - AH) \psi \rangle$.
So, we see the operator $HA - AH$ once again; note that if $A, H$ commute then the expectation of the state vector (or the standard deviation for that matter) does not evolve with time. This is certainly true for $H$ itself. Note: an operator that commutes with $H$ is sometimes called a “constant of motion” (think: “total energy of a system in classical mechanics).

Note also that $|\frac{d}{dt} A | = |\frac{i}{\hbar} \langle \psi, (HA - AH) \psi \rangle | \leq 2 \Delta A \Delta H$

If $A$ does NOT correspond with a constant of motion, then it is useful to define an evolution time $T_A = \frac{\Delta A}{\frac{E(A)}{dt}}$ where $\Delta A = (V(A))^{1/2}$ This gives an estimate of how much time must elapse before the state changes enough to equal the uncertainty in the observable.

Note: we can apply this to $H$ and $A$ to obtain $T_A \Delta H \ge \frac{\hbar}{2}$

Consequences: if $T_A$ is small (i. e., the state changes rapidly) then the uncertainty is large; hence energy is impossible to be well defined (as a numerical value). If the energy has low uncertainty then $T_A$ must be large; that is, the state is very slowly changing. This is called the time-energy uncertainty relation.