The following question is from David Griffiths' Introduction to Quantum Mechanics:
Problem 2.13 A particle in the harmonic oscillator potential starts out in the state
$$\Psi(x,0) = A[3 \psi_0(x) + 4 \psi_1(x)]\,.$$
(a) Find $A$.(b) Construct $\Psi(x,t)$ and $|\Psi(x,t)|^2$.
Part (b) asks to construct the general solution from the instance at time $t = 0$.
The strategy given in the book is to add the time dependence to each stationary state, forming a general solution:
To fit $\Psi(x,0)$ you write down the general linear combination of these solutions:
$$
\Psi(x,0)=\sum_{n=1}^\infty c_n\psi_n(x):\tag{2.16}
$$
the miracle is that you can *always match the specified initial state by appropriate choice of the constants $c_1,c_2,c_3,\ldots$. To construct $\Psi(x,t)$ you simply tack onto each term its characteristic time dependence, $e^{-iE_nt/\hbar}$:
$$
\Psi(x,t)=\sum_{n=1}^\infty c_n\psi_n(x) e^{-iE_nt/\hbar}=\sum_{n=1}^\infty c_n\Psi_n(x,t).\tag{2.17}
$$
I'm not sure, however, about how we can guarantee that two solutions $ {\psi_1}$ and $ {\psi_2}$ to the time-dependent equation don't have $ {\psi_1(x,0)} = {\psi_2(x,0)}$. If we can't guarantee this, then how do we know that the solution found by Griffiths' method is unique?
Best Answer
I interpret that your question basically asks how do we know that the time-dependent Schrödinger equation with an initial condition has a unique solution. To show this, we use a common method for showing the uniqueness of a solution to a linear differential equation: we show that if two answers are the same at one point, then their difference is constant $0$.
The time-dependent Schrödinger equation is linear; that means that if $\Psi_1$ and $\Psi_2$ are solutions to the time-dependent Schrödinger equation, then so is $\alpha_1 \Psi_1 + \alpha_2 \Psi_2$ for any $\alpha_1, \alpha_2$. In particular, $\Psi=\Psi_1-\Psi_2$ is a solution. So we have $$ -i\hbar \partial_t \Psi = \hat{H} \Psi,\quad\Psi(0)=0 $$ Now we get $$ \begin{align*} \partial_t \langle\Psi, \Psi\rangle &= \langle\partial_t \Psi, \Psi\rangle + \langle\Psi, \partial_t \Psi\rangle \\ &= \langle \frac{i}{\hbar} \hat{H} \Psi, \Psi \rangle + \langle \Psi, \frac{i}{\hbar} \hat{H} \Psi\rangle \\ = & -\frac{i}{\hbar}\langle\hat{H}\Psi,\Psi\rangle + \frac{i}{\hbar} \langle \Psi, \hat{H} \Psi\rangle \\ &= -\frac{i}{\hbar}\langle\Psi,\hat{H}\Psi\rangle + \frac{i}{\hbar} \langle \Psi, \hat{H} \Psi\rangle \\ &= 0, \end{align*} $$ where the next-to-last equation is due to the fact that the Hamiltonian is Hermitian. So now we have a differential equation for the real function $f$ defined by $f(t) \equiv \langle\Psi(t),\Psi(t)\rangle$, $$ \partial_t f = 0 \, . $$ Zero derivative means that $f(t)$ is a constant. Therefore, $$f(t)=f(0)=\left<\Psi(0),\Psi(0)\right>=0 \, .$$ Therefore, $\Psi=0$, so $\Psi_1=\Psi_2$.