Here is a quote from Introduction to quantum mechanics by David J Griffiths:

  1. 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?

  • $\begingroup$ To understand this you need some knowledge of Partial differential equations and Functional analysis (Hilbert's spectral theorem in particular). Linear Algebra is not, in general, enough. $\endgroup$
    – nabla
    Commented Feb 21, 2017 at 17:33
  • $\begingroup$ Related: physics.stackexchange.com/q/68822/2451 and links therein. $\endgroup$
    – Qmechanic
    Commented Feb 21, 2017 at 18:28

2 Answers 2


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 $E_1=E_2$.

  • $\begingroup$ I get it now, I was confusing the eigenvalues with coefficients. But as you said, eigenvalues don't enter the argument. They will only be considered if I express my wave functions in eigen basis. As a beginner in quantum mech, I don't even know if that is possible. $\endgroup$
    – sigsegv
    Commented Feb 22, 2017 at 12:38
  • $\begingroup$ I just added an extra bit to clarify some distinction with the time-independent solution. And yes: the Schrodinger operator has a complete set of eigenfunctions so any solution can be expressed as a sum of functions in this complete set. $\endgroup$ Commented Feb 22, 2017 at 13:48
  • $\begingroup$ Just want to add something that linearity gives you the idea that linear superposition of some specific solutions will be a solution too. But still, it doesn't tell you that the linear combination of such many solutions will be a general solution. That idea comes from completeness. $\endgroup$
    – Prabhat
    Commented Oct 17, 2020 at 9:16
  • $\begingroup$ @Prabhat ?? Completeness guarantees there's an expansion of any function in terms of solutions (assuming the correct boundary conditions), but linearity guarantees that a combination of however many solutions will be also be a solution. $\endgroup$ Commented Oct 17, 2020 at 18:06
  • $\begingroup$ @ZeroTheHero yes, you are right with both the definitions, but I was pointing out to the definition of 'general solution', you should call something a general solution when all the possible solutions can be expressed as that something. Talking particularly about Schrodinger's equation, what you get everytime after solving it, are eigenstates but it's completeness of eigenstate that we say the linear combination of them would be a general solution. I could be wrong or confusing things, otherwise. Will appreciate further discussion. $\endgroup$
    – Prabhat
    Commented Oct 21, 2020 at 4:10

Maybe to answer your question it is useful to start from a slightly different perspective, conceptually. In Quantum Mechanics a system is described by giving a Hilbert space, whose vectors represent states of the system (actually these are only part of the "states" called pure states, and any two vectors that are complex multiples of each other represent the same state. Don't worry about this for the moment) and a prescription of how the state of system evolves with time. In this case this prescription is the Schroedinger equation, which I write as an equation of Hilbert space vectors (in your case, the Hilbert space is the space of square integrable functions over $\mathbb{R}$, called $L^2(\mathbb{R})$) $$i\hbar\partial_t\psi_t=\hat{H}[\psi_t]$$

The operator $\hat{H}$ is self adjoint and mathematics tells us that in this case the unique solution to this equation with given initial condition $\psi_0$ (i.e. the state your system starts in) is $$\psi_t=e^{-\frac{i}{\hbar}\hat{H}t}[\psi_0]$$

where $U_t:=e^{-\frac{i}{\hbar}\hat{H}t}$ is a unitary operator called "time evolution" (compare this to the fact that the matrix exponential of $iA$, where $A$ is a hermitian matrix, is a unitary matrix).

Now the entire remaining work is to calculate $U_t\psi_0$ for given $\psi_0$. This is unfortunately really hard in most cases! That's where the time-independent Schroedinger equation comes in. Suppose you have a bunch of eigenvectors $\phi_1, \phi_2 \dots$ of $\hat{H}$, that is they satisfy $H\phi_i=E_i\phi_i$ (I hope there is no confusion when subscripts designate time and when they designate which $\phi$ I'm talking about). Again, mathematics tells us (and it's easy to verify this for matrices, that is for finite dimensional Hilbert spaces) that $$U_t[\phi_i]=e^{-\frac{i}{\hbar}\hat{H}t}[\phi_i]=e^{-\frac{i}{\hbar}E_it}\phi_i$$

This is really nice, because now we don't have to calculate the exponential of an operator. Also all these operators are linear, so if we manage to write our initial state $\psi_0$ as some linear combination of eigenvectors $$\psi_0=\sum_{i=1}^{?} c_i\phi_i$$ for some constants $c_i$ the desired solution takes the form

$$\psi_t=\sum_{i=1}^{?} c_ie^{-\frac{i}{\hbar}E_it}\phi_i$$ Thus we have solved the problem if we can find a way to express any initial condition we might want as a linear combination of eigenvectors of $\hat{H}$. We want to find an orthonormal basis of the Hilbert space consisting of such eigenvectors, then we can express ANY vector as an infinite linear combination.


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