I just started watching the coursera lectures on the basics of quantum mechanics and one of the first lectures were on deriving Schrodinger's equation and its interpretation it under Born's interpretation. What I want to ask is what the wave function \begin{equation} \psi({\bf r},t) \end{equation} return and represent. Now I know that \begin{equation} |\psi({\bf r},t)|^2dxdydz \end{equation} is the probability of finding the quantum particle described by \begin{equation} \psi({\bf r},t) \end{equation} in the volument element \begin{equation} dV = dxdydz \end{equation} at time T. But im not sure what the wave function $\psi$ returns. Could someone please explain in laymans term the return type of the $\psi$ function and what the lone $\psi$ function represents.

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    $\begingroup$ If you assume that probability is an inherent part of nature, then looking at this calculation books.google.ie/… we see probability (here it is the 'expected value') results in some vectors and matrices. We let matrix operators represent things like position, velocity and momentum (things we measure), & they operate on systems: applying the momentum operator on a vector is analogous to measuring the momentum of a system represented by that vector, thus the wave function (a vector) represents the state of a system. $\endgroup$
    – bolbteppa
    Aug 5, 2014 at 2:02

3 Answers 3


If this were computer science, we might say $\psi$ takes a $d$-tuple of reals ($r$) and another real ($t$) and returns a complex number with the attached unit of $L^{-d/2}$ in $d$ dimensions (with $L$ being the unit of length).1

If you want any more of an interpretation, well then you've already given it: $\psi(r,t)$ is the thing such that $\int_R\ \lvert \psi(r,t) \rvert^2\ \mathrm{d}V$ is the probability of the particle being observed in the region $R$ at time $t$. You can loosely think of it as a "square root" of a probability distribution.

The reason the "square root" interpretation is not quite right, and probably the reason you aren't satisfied with the $\int_R\ \lvert \psi(r,t) \rvert^2\ \mathrm{d}V$ definition, is that any particular instance of $\psi(r,t)$ carries extraneous information beyond what is needed to fully specify the physics. In particular, if we have $\psi_1$ describing a situation, then the wavefunction defined by $\psi_2(r,t) = \mathrm{e}^{i\phi} \psi_1(r,t)$ gives identical physics for any real phase $\phi$.

So the return value of the wavefunction itself is not a physical observable -- one always takes a square magnitude or does some other such thing that projects many mathematically distinct functions onto the same physical state. Even once you've taken the square magnitude, $\lvert \psi(r,t) \rvert^2$ arguably isn't directly observable, as all we can measure is $\int_R\ \lvert \psi(r,t) \rvert^2\ \mathrm{d}V$ (though admittedly for arbitrary regions $R$).

1You can check that $-d/2$ is necessarily the exponent. We need some unit such that squaring it and multiplying by the $d$-dimensional volume becomes a probability (i.e. is unitless). That is, we are solving $X^2 L^d = 1$, from which we conclude $X = L^{-d/2}$.

  • $\begingroup$ Great answer! especially for a programmer like myself :) Everything was super clear except for when you said "a complex number with the attached unit of L^−d/2 in d dimensions" could you please expand on that statement and explain why it is negative d-half's instead of just d. Thank you! $\endgroup$ Aug 5, 2014 at 2:32
  • $\begingroup$ @ArmenAghajanyan see footnote. Also note there's no rush to accept answers on this site; even better explanations may come in after a while. $\endgroup$
    – user10851
    Aug 5, 2014 at 2:38
  • $\begingroup$ Thank you for the advice, i am relatively new to this site. What do you mean by attached unit? $\endgroup$ Aug 5, 2014 at 2:40
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    $\begingroup$ Continuing with programming terminology: if you want to type every quantity you see in physics, specifying integer or real or complex is often not sufficient. There is a difference between a real mass quantity and a real velocity quantity; their product is a real momentum, not just a real, and their sum is not defined. Here $\psi$ returns a complex length^(-d/2) so to speak. $\endgroup$
    – user10851
    Aug 5, 2014 at 2:45
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    $\begingroup$ I'm not sure what $C^{-d/2}$ is meant to represent. There are quantities of type complex == complex length^0, complex length^(-1/2), complex length^(-1), etc. In 3D, $\psi$ returns a complex length^(-3/2), and the product of such a quantity with itself and 3 quantities of type real length is a dimensionless complex (and if you did everything right will turn out to be purely real). But perhaps I've taken the typing analogy too far... $\endgroup$
    – user10851
    Aug 5, 2014 at 2:57

The wavefunction returns a complex number whose modulus-squared is a probability density and whose phase is related to the probability current, i.e., where probability is flowing to.

If you write in the form $$\Psi({\mathbf x},t) = \sqrt{\rho({\mathbf x},t)}\exp\left(i\frac{S({\mathbf x},t)}{\hbar}\right)\text{,}$$ where $\rho(\mathbf{x},t)\geq0$ and $S(\mathbf{x},t)$ is real, then $\rho(\mathbf{x},t)$ is the probability density of measuring the position of the particle at $\mathbf{x}$ at time $t$.

Plugging this form into the Schrödinger equation with Hamiltonian $\hat{H} = -\frac{\hbar^2}{2m}\nabla^2 + V(\mathbf{x})$ gives: $$\begin{eqnarray*} \left[\frac{1}{2m}\left|\nabla S\right|^2 + V\right] + \frac{\partial S}{\partial t} &=& \frac{\hbar^2}{2m}\rho^{-1/2}\nabla^2\sqrt{\rho}\text{,}\\ \frac{\partial\rho}{\partial t} + \nabla\cdot\left(\frac{1}{m}\rho\nabla S\right) &=& 0\text{.} \end{eqnarray*}$$

The first equation is not important for our immediate purposes, but the second one is the continuity equation for the probability density $\rho$, forcing probability to be conserved. In other words, if we define $$\mathbf{J} = \frac{\rho}{m}\nabla S\text{,}$$ then where $\rho$ is the probability density, $\mathbf{J}$ is the probability current representing where the probability is flowing.

  • $\begingroup$ just to make sure were on the same page what is $ s(x,t) $ ? $\endgroup$ Aug 5, 2014 at 2:37
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    $\begingroup$ @ArmenAghajanyan: $S$ is the phase of the complex number that $\Psi$ returns multiplied by $\hbar$ ($S/\hbar$ is the phase itself). See Euler's formula to see that every complex number $z$ can be written in the form $z = re^{i\phi}$, where $r$ is the modulus and $\phi$ is the phase. This is the polar form of the complex number. $\endgroup$
    – Stan Liou
    Aug 5, 2014 at 2:41

I will answer in layman terms as from your age in your profile I would not expect a very strong background in the necessary mathematics.

When we have a function f(x) it returns a value at the point x, a real number. If it is the potential, V(r)=1/r we are able to calculate the potential and solve simple problems or enter the potential in complicated equations and solve complicated problems. This is a real function.

One can extend functions using complex numbers. Complex number functions are in reality two functions:

f1(x,y,z,t) + f2(x,y,z,t)*i

Here i is the square root of the real number (-1). Using complex numbers simplifies the form of equations generally and allows for easier manipulation of theoretical quantities.

The Schrodinger equation is an equation in complex numbers and thus

\begin{equation} \psi(r,t) \end{equation}

is a solution of the equation in our everyday space and time with two functions attached, it returns a complex value with two real numbers, the second one attached to i .

Our measurements in real life return real numbers, so as it is Psi has no use. That is why it is squared with its complex conjugate , to give a real number which will represent the probability in the volume element. This last is a postulate, the Born rule, and it has been found to work and has not been falsified experimentally, and thus we accept the quantum mechanical framework of nature.

  • $\begingroup$ A bit of a over simplification but none the less a good answer! So for my background in math i have done basic calculus, first order/second order differential equations, path integrals, worked extensively with the Laplace transform to solve differential equations. On the linear algebra stuff, apart from the basics i have studied both eigen and jordon decompositions. I have not yet got around to working on complex analysis, and have done most frequency analysis (fourier) only on the computational/coding side. Could you please recommend what type of math i should pursue? $\endgroup$ Aug 5, 2014 at 6:46
  • $\begingroup$ Sorry Armen, but if you look at my profile my math days were over by 1963. A lot more books then the ones I know have come up since then and most probably more appropriate to physics. Better ask Chris White. $\endgroup$
    – anna v
    Aug 5, 2014 at 12:04

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