While working out the stationary states of a single particle in a 3d infinite potential box ($V=0$ inside a cuboid of known dimensions, $V=\infty$ everywhere else), I realized I had to assume the wavefunction was separable into a product of three functions, $\psi(x,y,z)=X(x)Y(y)Z(z)$, in order to find $\psi$. Why is this so, and under what conditions? What guarantees me that I can do this? The text I'm following isn't particularly clear on this.

  • $\begingroup$ There are actually good rigorous proofs of the fact that the assumption of separability for particular systems does not "destroy" any information about the problem, and I vaguely recall they involve examining the invariance of the Hamiltonian under the action of Lie groups associated with the symmetry group of the system, but I honestly don't know the specifics. Needless to say, most intro books avoid mention of such difficulties entirely (as otherwise the text would spiral down a complicated path that would be pedagogically suicidal). $\endgroup$ Apr 23, 2014 at 22:39
  • $\begingroup$ I see. I was wondering if we could simply assume this to be true for nearly all practical applications, or if there were pathological counterexamples, or no counterexamples at all. $\endgroup$
    – andrepd
    Apr 23, 2014 at 22:45
  • $\begingroup$ There are many counterexamples. For example, you could try to assume that the particle in 2D box has eigenstates which are separable in polar coordinates $(r,\phi)$, but this fails because the eigenstates turn out to be not separable in those coordinates. However, most prototypical exactly solvable quantum systems are modeled in a coordinate system in which the system winds up being separable, so when you learn about them, you don't have to worry about it. But in general, it is not necessarily safe to assume the system is separable in any arbitrary coordinate system. $\endgroup$ Apr 23, 2014 at 23:16
  • 1
    $\begingroup$ @luming: No, that doesn't work at all. For example, the ground state $n_x=1,n_y=1$ of the 2D infinite square well box with $L_x=L_y=1$ is $2 \sin (\pi x) \sin (\pi y)$, which can not be written in the form $A(r)B(\theta)$. You can prove the impossibility by inserting $x\rightarrow r\cos(\theta),y\rightarrow r\sin(\theta)$ and then plotting the result for fixed $r$ and variable $\theta$. It will never be constant for any choice of $r>0$. Thus, separation of variables fails completely in that case, no matter how hard you try. $\endgroup$ Apr 24, 2014 at 1:26
  • 1
    $\begingroup$ @luming: Actually, forget about my reasoning in the sentence "plotting the result for fixed $r$ and variable $\theta$...", that was wrong. The correct reasoning is that in polar coordinates the ground state is given by $\psi(r,\theta)=\sin (\pi r \sin (\theta )) \sin (\pi r \cos (\theta ))$. However, the ratio $\psi(r_1,\theta)/\psi(r_2,\theta)$ is not independent of $\theta$ whenever $r_1\neq r_2$. However, if it were separable, you would obtain $A(r_1)/A(r_2)$, which is constant. This is a contradiction. $\endgroup$ Apr 24, 2014 at 2:09

2 Answers 2


Sepration of variables is indeed a delicate topic in partial differential equations. As of today we don't (to the extend of my knowledge) have a complete theory on the conditions that make it possible. The usual posture is to have existence and uniqueness theorems for the solutions of a given PDE and using some ansatz from separation of variables, by finding a general solution, we should have the solution as luming commented.

As far as I know, for specific cases we do have rigorous justification on using separation of variables in given coordinates, which are related to the symmetry group acting on the PDE (as BumbsterDoofus said also in comments). A (somewhat old) book explaining this is Miller's "Symmetry and Separation of Variables" that you can find online here http://www.ima.umn.edu/~miller/separationofvariables.html. As it says in the preface we do know how to justify for some PDEs (specially lower dimensional ones) but we do not have complete theory for all the differential equations we would like to consider (for instance the tridimensional wave equation). I do not know of further developments beyond Miller's book, but I have looked for it and have not found decisive changes (but that could be due to my ignorance).

In any case, as long as you're considering bound states I do not think you should be worried about those things, the existence and uniqueness theorems coupled with you being able to provide a general solution should suffice (I'm always suspicious of scattering states becaude they are not square-integrable and could be more subtle). If you're not satisfied with this answer, I guess it would make a great question on math stackexchange to ask for the status of separation of variables, though I think the answer relates to the symmetry group of the PDE in question anyway, and could be overkill for your context.


The logic goes like following.

We can guess solution in forms of $X(x) Y(y) Z(z)$ for a particle in 3-dimensional box. We can find such solutions. The question is, do we miss any solution?

The function $X(x)$ is eigenfunction of self-adjoint operator $$H_x = -\frac{1}{2} \frac{ \partial^2}{\partial x^2} + V(x) \tag{1} $$ $V(x)$ is the potential of infinity wall. And the same for $y,z$. Thus they form a complete set of functions under suitable boundary condition. We expand the general form of solution as $$ \psi(x,y,z) = \sum_{l} \sum_{m} \sum_{n} c_{lmn} X_l(x) Y_m(y) Z_n(z) \tag{2} $$

Since $[H_x,H]=0$, the eigenfunction of a particle in 3-dimensional box forms a simultaneous eigenstate as eigenfunction of $H_x$. We can drop $\sum_l$ and the $l$ dependence in the expansion coefficient $c_{clm}$ in Eq. (2). The same applies to $y$ and $z$. Therefore the eigenfunction of particle in three-dimensional box can be writen as $X(x) Y(y) Z(z)$ .


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.