# College Math Teaching

## July 28, 2011

### Quantum Mechanics and Undergraduate Mathematics VII: Time Evolution of the State Vector

Filed under: advanced mathematics, applied mathematics, physics, quantum mechanics, science — collegemathteaching @ 2:38 pm

Of course the state vector $\psi$ changes with time. The question is how does it change with time and how does the probability density function associated with an observable change with time?

Note: we will write $\psi_t$ for $\psi(x,t)$. Now let $A$ be an observable. Note that the eigenvectors and the eigenvalues associated with $A$ do NOT change with time, so if we expand $\psi_t$ in terms of the eigenbasis for $A$ we have $\psi_t = \sum_k \langle \alpha_k, \psi_t \rangle \alpha_k$ hence $\frac{\partial \psi_t}{\partial t} = \sum_k \langle \alpha_k, \frac{\partial \psi_t}{\partial t} \rangle \alpha_k$

Of course, we need the state vector to “stay in the class of state vectors” when it evolves with respect to time, which means that the norm cannot change; or $\frac{d}{dt} \langle \psi_t, \psi_t \rangle = 0$.

Needless to say there has to be some restriction on how the state vector can change with time. So we have another postulate:

Postulate 5
For every physical system there exists a linear Hermitian operator $H$ called the Hamiltonian operator such that:
1. $i\hbar \frac{\partial}{\partial t} \psi(x,t) = H\psi(x,t)$ and
2. $H$ corresponds to the total energy of the system and possesses a complete set of eigenvectors $\eta_k$ and eigenvalues $e_k$ where the eigenvalues are the “allowed values” of the total energy of the system.

Note: $\hbar$ is the constant $\frac{h}{\pi}$ where $h$ is Plank’s constant.

Note: $H$ is not specified; it is something that the physicists have to come up with by observing the system. That is, there is a partial differential equation to be solved!

But does this give us what we want, at least in terms of $\psi_t$ staying at unit norm for all times $t$? (note: again we write $\psi_t$ for $\psi(x,t)$ ).

The answer is yes; first note that $\frac{d}{dt} \langle \psi_t, \psi_t \rangle = \langle \frac{\partial \psi_t}{\partial t}, \psi_t \rangle + \langle \psi_t, \frac{\partial \psi_t}{\partial t}\rangle$; this is an easy exercise in using the definition of our inner product and differentiating under the integral sign and noting that the partial derivative operation and the conjugate operation commute).

Now note: $\langle \frac{\partial \psi_t}{\partial t}, \psi_t \rangle + \langle \psi_t, \frac{\partial \psi_t}{\partial t}\rangle = \overline{-\frac{i}{\hbar}}\langle H\psi_t, \psi_t \rangle + -\frac{i}{\hbar}\langle \psi_t, H \psi_t \rangle = \frac{i}{\hbar}(\langle H\psi_t, \psi_t \rangle - \langle \psi_t, H\psi_t \rangle) = 0$
because $H$ is Hermitian.

Note: at this point Gillespie takes an aside and notes that if one denotes the state vector at time $t = 0$ by $\psi_0$ then one can attempt to find an operator $U(t)$ where $\psi_t = U(t)\psi_0$. This leads to $i \hbar \frac{\partial U(t)}{\partial t} = HU\psi_0$ which must be true for all $\psi_0$. This leads to $i\hbar \frac{\partial U(t)}{\partial t} = HU(t)$ with initial condition $U(0) = 1$. This leads to the solution $U(t) = exp(\frac{-iHt}{\hbar})$ where the exponential is defined in the sense of linear algebra (use the power series expansion of $exp(A)$). Note: $U$ is not Hermitian and therefore NOT an observable. But it has a special place in quantum mechanics and is called the time-evolution operator.

Next: we’ll deal with the time energy relation for the expected value of a specific observable.