Quantum Mechanics and Undergraduate Mathematics IX: Time evolution of an Observable Density Function

August 9, 2011

Quantum Mechanics and Undergraduate Mathematics IX: Time evolution of an Observable Density Function

Filed under: advanced mathematics, applied mathematics, calculus, linear albegra, probability, quantum mechanics, science, series — collegemathteaching @ 7:43 pm

We’ll assume a state function $\psi$ and an observable whose Hermitian operator is denoted by $A$ with eigenvectors $\alpha_k$ and eigenvalues $a_k$ . If we take an observation (say, at time $t = 0$ ) we obtain the probability density function $p(Y = a_k) = | \langle \alpha_k, \psi \rangle |^2$ (we make the assumption that there is only one eigenvector per eigenvalue).

We saw how the expectation (the expected value of the associated density function) changes with time. What about the time evolution of the density function itself?

Since $\langle \alpha_k, \psi \rangle$ completely determines the density function and because $\psi$ can be expanded as $\psi = \sum_{k=1} \langle \alpha_k, \psi \rangle \alpha_k$ it make sense to determine $\frac{d}{dt} \langle \alpha_k, \psi \rangle$ . Note that the eigenvectors $\alpha_k$ and eigenvalues $a_k$ do not change with time and therefore can be regarded as constants.

$\frac{d}{dt} \langle \alpha_k, \psi \rangle = \langle \alpha_k, \frac{\partial}{\partial t}\psi \rangle = \langle \alpha_k, \frac{-i}{\hbar}H\psi \rangle = \frac{-i}{\hbar}\langle \alpha_k, H\psi \rangle$

We can take this further: we now write $H\psi = H\sum_j \langle \alpha_j, \psi \rangle \alpha_j = \sum_j \langle \alpha_j, \psi \rangle H \alpha_j$ We now substitute into the previous equation to obtain:
$\frac{d}{dt} \langle \alpha_k, \psi \rangle = \frac{-i}{\hbar}\langle \alpha_k, \sum_j \langle \alpha_j, \psi \rangle H \alpha_j \rangle = \frac{-i}{\hbar}\sum_j \langle \alpha_k, H\alpha_j \rangle \langle \alpha_j, \psi \rangle$

Denote $\langle \alpha_j, \psi \rangle$ by $a_j$ . Then we see that we have the infinite coupled differential equations: $\frac{d}{dt} a_k = \frac{-i}{\hbar} \sum_j a_j \langle \alpha_k, H\alpha_j \rangle$ . That is, the rate of change of one of the $a_k$ depends on all of the $a_j$ which really isn’t a surprise.

We can see this another way: because we have a density function, $\sum_j |\langle \alpha_j, \psi \rangle |^2 =1$ . Now rewrite: $\sum_j |\langle \alpha_j, \psi \rangle |^2 = \sum_j \langle \alpha_j, \psi \rangle \overline{\langle \alpha_j, \psi \rangle } = \sum_j a_j \overline{ a_j} = 1$ . Now differentiate with respect to $t$ and use the product rule: $\sum_j \frac{d}{dt}a_j \overline{ a_j} + a_j \frac{d}{dt} \overline{ a_j} = 0$

Things get a bit easier if the original operator $A$ is compatible with the Hamiltonian $H$ ; in this case the operators share common eigenvectors. We denote the eigenvectors for $H$ by $\eta$ and then
$\frac{d}{dt} a_k = \frac{-i}{\hbar} \sum_j a_j \langle \alpha_k, H\alpha_j \rangle$ becomes:
$\frac{d}{dt} \langle \eta_j, \psi \rangle = \frac{-i}{\hbar} \sum_j \langle \eta_j, \psi \rangle \langle \eta_k, H\eta_j \rangle$ Now use the fact that the $\eta_j$ are eigenvectors for $H$ and are orthogonal to each other to obtain:
$\frac{d}{dt} \langle \eta_k, \psi \rangle = \frac{-i}{\hbar} e_k \langle \eta_k, \psi \rangle$ where $e_k$ is the eigenvalue for $H$ associated with $\eta_k$ .

Now we use differential equations (along with existence and uniqueness conditions) to obtain:
$\langle \eta_k, \psi \rangle = \langle_k, \psi_0 \rangle exp(-ie_k \frac{t}{\hbar})$ where $\psi_0$ is the initial state vector (before it had time to evolve).

This has two immediate consequences:

1. $\psi(x,t) = \sum_j \langle \eta_j, \psi_0 \rangle exp(-ie_j \frac{t}{\hbar}) \eta_j$
That is the general solution to the time-evolution equation. The reader might be reminded that $exp(ib) = cos(b) + i sin (b)$

2. Returning to the probability distribution: $P(Y = e_k) = |\langle \eta_k, \psi \rangle |^2 = |\langle \eta_k, \psi_0 \rangle |^2 ||exp(-ie_k \frac{t}{\hbar})|^2 = |\langle \eta_k, \psi_0 \rangle |^2$ . But since $A$ is compatible with $H$ , we have the same eigenvectors, hence we see that the probability density function does not change AT ALL. So such an observable really is a “constant of motion”.

Stationary States
Since $H$ is an observable, we can always write $\psi(x,t) = \sum_j \langle \eta_j, \psi(x,t) \rangle \eta_j$ . Then we have $\psi(x,t)= \sum_j \langle \eta_j, \psi_0 \rangle exp(-ie_j \frac{t}{\hbar}) \eta_j$

Now suppose $\psi_0$ is precisely one of the eigenvectors for the Hamiltonian; say $\psi_0 = \eta_k$ for some $k$ . Then:

1. $\psi_(x,t) = exp(-ie_k \frac{t}{\hbar}) \eta_k$
2. For any $t \geq 0 , P(Y = e_k) = 1, P(Y \neq e_k) = 0$

Note: no other operator has made an appearance.
Now recall our first postulate: states are determined only up to scalar multiples of unity modulus. Hence the state undergoes NO time evolution, no matter what observable is being observed.

We can see this directly: let $A$ be an operator corresponding to any observable. Then $\langle \alpha_k, A \psi_k \rangle = \langle \alpha_k, A exp(-i e_k \frac{t}{\hbar})\eta_k \rangle = exp(-i e_k \frac{t}{\hbar}\langle \alpha_k, A \eta_k \rangle$ . Then because the probability distribution is completely determined by the eigenvalues $e_k$ and $|\langle \alpha_k, A \eta_k \rangle |$ and $|exp(-i e_k \frac{t}{\hbar}| = 1$ , the distribution does NOT change with time. This motivates us to define the stationary states of a system: $\psi_{(k)} = exp(- e_k \frac{t}{\hbar})\eta_k$ .

Gillespie notes that much of the problem solving in quantum mechanics is solving the Eigenvalue problem: $H \eta_k = e_k \eta_k$ which is often difficult to do. But if one can do that, one can determine the stationary states of the system.

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College Math Teaching

August 9, 2011