Stationary States

Stationary states are states for which the probability density $\rho(x,t)=\psi^*(x,t)\psi(x,t)$ is constant at all times (i.e., states for which $\frac{\partial{\rho(x,t)}}{\partial{t}} = 0$, and therefore $\nabla \cdot {\bf j} = 0$). In this section we will show that if $\psi(x, t)$ is factorizable according to $\psi(x,t) = \phi(x) f(t)$, then $\psi(x, t)$ is a stationary state.

Substituting $\psi(x, t)$ in the time dependent Schrödinger equation we obtain:

$$\phi(x) i\hbar \frac {\partial{f(t)}}{\partial{t}} = -f(t)\frac{\hbar^2}{2m} \frac{\partial^2\phi(x)}{\partial x^2} +f(t)V(x)\phi(x),$$

and dividing both sides by $f(t)\phi(x)$ we obtain:

$$\frac{i\hbar}{f(t)}\frac{\partial{f(t)}}{\partial{t}}=-\frac{\hbar^2}{2m\phi(x)}\frac{\partial^2\phi(x)}{\partial x^2}+V(x).$$ (4)

Since the right hand side (r.h.s) of Eq. (4) can only be a function of $x$ and the l.h.s. can only be a function of $t$ for any $x$ and $t$, and both functions have to be equal to each other, then such function must be equal to a constant E. Mathematically,

$$\frac{i\hbar}{f(t)} \frac{\partial f(t)}{\partial {t}} =E \Rightarrow f(t)=f(0)$$   exp$$(-\frac{i}{\hbar}Et),$$

$$-\frac{\hbar^2}{2m\phi(x)} \frac{\partial^2 \phi(x)}{\partial{x^2}} +V(x) = E \Rightarrow \boxed{\hat{H}\phi(x)=E\phi(x)}.$$

The boxed equation is called the time independent Schrödinger equation.

Furthermore, since $f(0)$ is a constant, function $\tilde{\phi}(x)=f(0)\phi(x)$ also satisfies the time independent Schrödinger equation as follows,

$$\boxed{\hat{H}\tilde{\phi}(x)=E\tilde{\phi}(x)},$$ (5)

and

$$\psi(x, t) = \tilde{\phi}(x)$$   exp$$(-\frac{i}{\hbar}Et).$$

Eq. (5) indicates that $E$ is the eigenvalue of $\hat{H}$ associated with the eigenfunction $\tilde{\phi}(x)$.

Exercise 3: Prove that $\hat{H}$ is a Hermitian operator.

Exercise 4: Prove that -i $\hbar \partial / \partial{x}$ is a Hermitian operator.

Exercise 5: Prove that if two hermitian operators $\hat{Q}$ and $\hat{P}$ satisfy the equation $\hat{Q} \hat{P} = \hat{P} \hat{Q}$, i.e., if $P$ and $Q$ commute (vide infra), the product operator $\hat{Q} \hat{P}$ is also hermitian.

Since $\hat{H}$ is hermitian, $E$ is a real number $\Rightarrow E=E^*$ (see Property 1 of Hermitian operators), then,

$$\psi^*(x,t)\psi(x,t)=\tilde{\phi}^*(x) \tilde{\phi}(x).$$

Since $\tilde{\phi}(x)$ depends only on $x$, $\frac{ \partial }{\partial{t}} (\tilde{\phi}^*(x) \tilde{\phi}(x)) = 0$, then, $\frac{ \partial }{\partial{t}}\psi^*(x,t) \psi(x,t) = 0$. This demonstration proves that if $\psi(x,t) = \phi(x) f(t)$, then $\psi(x, t)$ is a stationary function.