Here is a quote from Introduction to quantum mechanics by David J Griffiths:
- The general solution is a linear combination of separable solutions. As we're about to discover, the time-independent Schroedinger equation (Equation 2.5) yields an infinite collection of solutions ($\psi_1(x)$, $\psi_2(x)$, $\psi_3(x)$,…), each with its associated value of the separation constant ($E_1$, $E_2$, $E_3$,…); thus there is a different wave function for each allowed energy:
$$\Psi_1(x, y) = \psi_1(x)e^{-iE_1 t/\hbar},\quad \Psi_2(x, y) = \psi_2(x)e^{-iE_2 t/\hbar}, \ldots.$$
Now (as you can easily check for yourself) the (time-dependent) Schroedinger equation (Equation 2.1) has the property that any linear combination5 of solutions is itself a solution. Once we have found the separable solutions, then, we can immediately construct a much more general solution, of the form
$$\Psi(x, t) = \sum_{n = 1}^{\infty}c_n\psi_n(x)e^{-iE_n t/\hbar}\tag{2.15}$$
I am trying to understand it in this way.
…the time independent Schroedinger's equation $\hat H\psi = E\psi$
An eigenvalue equation $Ax = \lambda x$,
yields an infinite collection of solutions ($\psi_1(x)$, $\psi_2(x)$, $\psi_3(x)$, $\dots$)
has eigen vectors $x_1$, $x_2$, $x_3$, $\dots$
each with it's associated value of separation constant ($E_1$, $E_2$, $E_3$, $\dots$);
each with it's associated eigen value $\lambda_1$, $\lambda_2$, $\lambda_3$, $\dots$
thus there is a different wave function for allowed energy:
$$\Psi_1(x,t) = \psi_1(x)e^{-iE_1t/\hbar},\quad\Psi_2(x,t) = \psi_2(x)e^{-iE_2t/\hbar}, \dots$$
have equations as
$$Ax_1=\lambda_1x_1, \qquad Ax_2=\lambda_2x_2, \dots$$
Once we have found the separable solutions, then, we can
immediately construct a much more general solution, of the form
$$\Psi(x,t) = \sum_{n=1}^{\infty}c_n\psi_n(x)e^{-E_nt/\hbar}$$
(Forgetting the any other variable dependence) We can construct a more general solution of the form $$X = \sum_{n}c_n x_n$$
This last equation doesn't make any sense to me. There is nothing in linear algebra that says that this last equation logically precedes the previous equations. Trying to understand from linear algebra, what does the last equation mean? Why is the general solution of Schroedinger's equation a linear combination of the eigenfunctions?
Best Answer
You are starting from the incorrect point. The argument follows by linearity of the equation.
Suppose $\Psi_k(x,t)$ is solution of the time dependent Schr$\ddot{\hbox{o}}$dinger equation: $$ i\hbar \frac{\partial }{\partial t}\Psi_k(x,t)=-\frac{\hbar^2}{2m}\frac{\partial^2\Psi_k(x,t)}{\partial x^2}+U(x)\Psi_k(x,t)\, . $$ Then: $$ \Phi(x,t)=a_1\Psi_1(x,t)+a_2\Psi_2(x,t) $$ is also a solution since $$ i\hbar \frac{\partial }{\partial t}\Phi(x,t) =a_1\left(i\hbar \frac{\partial }{\partial t}\Psi_1(x,t)\right)+a_2 \left(i\hbar \frac{\partial }{\partial t}\Psi_2(x,t)\right) $$ and \begin{align} -\frac{\hbar^2}{2m}\frac{\partial^2\Phi(x,t)}{\partial x^2}+U(x)\Phi(x,t) &=a_1\left(-\frac{\hbar^2}{2m}\frac{\partial^2\Psi_1(x,t)}{\partial x^2}+U(x)\Psi_1(x,t)\right)\\ &\quad + a_2\left(-\frac{\hbar^2}{2m}\frac{\partial^2\Psi_2(x,t)}{\partial x^2}+U(x)\Psi_2(x,t)\right)\, . \end{align} These follow simply from the known rule valid for any two differentiable functions $f$ and $g$: $\partial (f+g)/\partial t=\partial f/\partial t+\partial g/\partial t$, and similarly for the partials w/r to $x$. Combining these last two equations you get an identity for any $a_1$ and $a_2$ since each $\Psi_k(x,t)$ is independently a solution. Of course this simply extends to an arbitrary number of terms in the linear combination.
Note the eigenvalue of the time-independent part never enters in this argument. The final step is to observe that separation of variables in the time-dependent equation yields $\Psi_k(x,t)=e^{-iE_k t}\psi_k(x)$ with $\psi_k(x)$ an eigenfunction of the time-independent equation, but again, this does not enter in the argument.
Edit: note this is in contradistinction with the time-independent equation. When $$ -\frac{\hbar^2}{2m}\frac{d^2\psi_k(x)}{dx^2}+U(x)\psi_k(x)=E_k\psi_k(x) $$ the the right hand side is must a multiple of the original function. With this observation, note then that a linear combination $$ \psi(x)=a_1\psi_1(x)+a_2\psi_2(x) $$ will in general NOT be a solution of the time-independent equation because \begin{align} \left(-\frac{\hbar^2}{2m}\frac{d^2}{dx^2}+U(x)\right)\psi(x) &=a_1\left(-\frac{\hbar^2}{2m}\frac{d^2}{dx^2}+U(x)\right)\psi_1(x)\\ &\qquad+a_2\left(-\frac{\hbar^2}{2m}\frac{d^2}{dx^2}+U(x)\right)\psi_2(x) \\ &=a_1E_1\psi_1(x)+a_2E_2\psi_2(x)\\ &=E_1(a_1\psi_1(x)+a_2\psi_2(x))+(E_2-E_1)a_2\psi_2(x)\\ &=E_1\psi(x)+(E_2-E_1)a_2\psi_2(x) \end{align} will NOT be a multiple of $\psi(x)$ unless $E_1=E_2$.