I have just started Griffiths Intro to QM. I was studying Born's interpretation of Wave function and it says that the square of the modulus of the wave function is a measure of the probability of finding the particle at that position. But if wave function is complex then surely its modulus would be real and non negative.

  1. If so then why should we square the modulus when in reality modulus itself should give real and non negative values which are equivalent for probability?

  2. Does this squaring have anything to do with the fact that the Schrodinger equation is of second order?

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    $\begingroup$ @Jimself the modulus of a complex number $z$ is $\sqrt{\bar{z}z}$ and is always positive... $\endgroup$
    – yuggib
    Commented Jul 20, 2015 at 14:18
  • $\begingroup$ Thank you @Jimself for the reply. But a complex number is represented using only modulus which is non negative. Isnt it so ? Modulus in essence is the distance of the point from the origin of the complex plane ? how is it negative ? I am confused. $\endgroup$ Commented Jul 20, 2015 at 14:21
  • $\begingroup$ @yuggib right, I was just temporarily sideblinded wondering what it being complex had to do with it $\endgroup$
    – Jim
    Commented Jul 20, 2015 at 14:26

4 Answers 4


Well there are multiple reasons, but a very important one is that it can be proven (from the Schrödinger equation) that

$$\frac{\mathrm d}{\mathrm dt}\int \mathrm d\boldsymbol x\ |\psi(\boldsymbol x,t)|^2=0$$

so that, if at any moment in time we have $\int \mathrm d\boldsymbol x\ |\psi(\boldsymbol x,t)|^2=1$, this will remain true at any other time.

On the other hand, the derivative of the integral of $|\psi|$ is not time independent, so a consistent normalization is not possible

We need that the integral to be time independent, because otherwise a probabilistic interpretation wouldn't be possible. We need that the probability of finding the particle somewhere has to be $1$. If we used $|\psi|$ as a probability distribution, and at any point in time had we that the integral equals $1$, this will change over time, what wouldn't make any sense. On the other hand, as I already stated, we can think of $|\psi|^2$ as a probability just because its integral is time-independent. So if at any point in time we have that the integral of $|\psi|^2$ equals $1$, this will remain true at any other point in time.

Also, this has nothing to do with the Schrödinger equation being of 2nd order: Dirac equation is a 1st order equation and (in some sense), the probability distribution is still $\psi^\dagger\psi$.

Edit: there is another explanation that might be more "physical", closer to our intuition. You probably know about the double slit-experiment, a standard way of introducing QM. When learning about such experiment, we are given two scenarios: first, think of the double slit being hit by light. We know from optics about the phenomenon of interference: the electromagnetic field is radiated from each slit, thus interfering when reaching the screen. The interference pattern is easily understood, mathematically, when we think of the electric field as a wave propagating through space. We know that the intensity observed at the screen is the modulus squared of $\boldsymbol E$, where $\boldsymbol E=\boldsymbol E_1+\boldsymbol E_2$. When calculating the modulus squared, we get the expected interference (crossed) term. The observed intensity is just $I(x)=|\boldsymbol E(x)|^2$.

On the other hand, if we think of the experiment when using electrons, we know that the interference pattern is still produced, so by being inspired from classical electrodynamics, we think of another wave propagating through space, such that its modulus squared gives the intensity on the screen, i.e., the modulus squared of the wave function is like the intensity of the light: where it is high, there is a high chance of finding an electron. In this way, we can think of $|\psi|^2$ as a probability distribution, in the same way we can think of $|\boldsymbol E|^2$ as a probability distribution of the photon. There is actually a lot from QM taken from classical electromagnetism.

For the record, I must say that this analogy between the electric field and the wave-function is rather limited, and should not be pushed too far: it will lead to incorrect conclusions. The electric field is not the wave function of the photon.

  • $\begingroup$ I should add that actually Dirac's equation is more complicated (and subtle) than Schrodinger's, for many reasons (note that I wrote $\psi^\dagger$ instead of $\psi^*$); in Dirac's case, $\psi$ is not a really a function, and $\psi^\dagger\psi$ is not really a probability distribution, but it is related somehow. $\endgroup$ Commented Jul 20, 2015 at 14:34
  • $\begingroup$ thanks @qftishard. I think I get the idea now. So it has all to do with time independence and the fact that probability has to be 1 at some position. And such a feature will not appear if we use just the modulus. Thanks. $\endgroup$ Commented Jul 20, 2015 at 14:35
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    $\begingroup$ (2/2) You will soon meet the concept of an operator through your journy with QM. You got to be patient: it is a beautiful journy, but certainly not an easy one. Stay there, just don't give up :) $\endgroup$ Commented Jul 20, 2015 at 18:05
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    $\begingroup$ Thanks for providing the link to that question. +1 for The electric field is not the wave function of the photon. $\endgroup$
    – user36790
    Commented Nov 1, 2015 at 11:28
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    $\begingroup$ I must be missing something obvious—why is the (derivative of the integral of the) mod squared time independent but just the modulus is not? $\endgroup$
    – T3db0t
    Commented May 19, 2018 at 0:01

Your suspicion that the square of the modulus of the wave function has to be used as the probability density is indeed related with the Schrodinger equation. I approach the question by using eigen value equation theory. The Schrodinger equation is a eigenvalue equation. The wave function is a superposition of eigenfunctions of the Schrodinger equation. Or different words, the wave function can be developed into a complete set of eigen functions : $\Psi = \sum_l a_l\psi_l$ each of which fulfills the Schrodinger equation corresponding to a particular energy eigen value $E_l$: $H\psi_l =E_n\psi_l$. The coefficients $a_l$ express the probability (actually it is $|a_l|^2$) to find a particle with wave function $\Psi$ in state $\psi_l$.

The different $\psi_l$ have to be orthogonal to each other, not only due to mathematical reasons, also to a physical one. If an energy measurement is carried out and the it's found that the particle governed by the Schrodinger equation is found to be in energy state $E_n$ then it will stay in that energy state. That means after the measurement $\Psi =\psi_n$ and $a_l= \delta_{ln}$ (The wave function $\Psi$ collapses to a particular function $\psi_l$.). It is expressed in the quantum mechanical formalism as that the projection of the wave function $\psi$ on another energy state eigen function $m\ \neq n$ $\int \Psi^{\star} \psi_m dx =0$ whereas the projection of the wavefunction on the eigen function of state $n$ is one: $\int \Psi^{\star} \psi_n dx =1$. You can check this by putting $\Psi = \sum_l a_l\psi_l$ with $a_l= \delta_{ln}$ in the integral. The eigen functions of the Schrodinger eigen value equation fulfill very well this property (the eigen function can of course easily be normalized so that they fulfill $\int \psi_m^{\star} \psi_m dx =1$ for all $m$). From this we learn that in a particular state $m$ $\int |\psi_m|^2 dx =1$. The same consideration can be made for any other well defined operator and its corresponding eigen states. Finally as a physicist you would like to make sense of this relation $\int |\psi_m|^2 dx =1$. And Born ended up stating that it means that the $|\psi_m|^2$ is the probability density to find a particle in the interval $dx$. $x$ does not need to necessarily mean position, a priori it could also be momentum $p$ etc. But the concept remains the same: it would mean that $|\psi_m|^2$ is the probability density to find the particle in momentum interval $dp$. And obviously trying to define the probability density with $\int |\psi_m| dx =1$ does not make much sense as this expression $\int |\psi_m| dx =1$ does not appear in the eigen mode formalism. It is may be useful to stress that eigen value equations like the Schrodinger equation appear everywhere in physics for instance in electrodynamics and the eigen functions fulfill the same relations as those of the Schrodinger equation. This means the formalism does not fall from the sky, it's also used in classical physics. It the way to solve eigenvalue equations in mathematical physics. But on top of it a typical quantum mechanical assumption is used: The collapse of the wave function upon a measurement to a particular wave function. And finally the orthogonality relations get a particular physical, even quantum-mechanical sense.

My explanation is certainly not the most rigorous one, in full-fleshed QM-formalism with bras and ket vectors it can be shown more rigorously. And I recommend you to continue to read the book, although I don't know it it will certainly demonstrate this in some way in the following.

  • $\begingroup$ Thank you @Frederic Thomas. I must confess this answer was a bit difficult for me to understand. Like I said, I am a beginner and I am studying on my own. But on reading it twice, I think I get some hold of it. :) $\endgroup$ Commented Jul 20, 2015 at 16:07
  • $\begingroup$ @anandtr2006: yeah, finally I've found it not very convincing. The answer from qftishard is probably more conclusive. The only thing I wanted to say is that it is rather natural to work with products of wave functions including squared moduli instead of moduli $\endgroup$ Commented Jul 23, 2015 at 15:10
  • $\begingroup$ Thank you Frederic Thomas, But I appreciate your effort. I have a lot of work to catch upon, I guess. :D $\endgroup$ Commented Jul 25, 2015 at 3:10

The interpretation of $|\psi|^2$ as probability of a particle in given region comes from optics. In optics, pattern of interference has dark and bright fringes, but detection of fringes is scalar quantity that is intensity of light in given region of width. The intensity of wave is given as square of its amplitude and angular frequency. But interference is spatial phenomena or time-independent, as dark fringe remain dark at any value of time, so it was considered that intensity is proportional to square of amplitude.

Now quantum mechanics is wave mechanics, mathematics is similar but interpretation varies. But particles and waves are two exclusive phenomenon, to join them De broglie's hypothesis was used, matter wave. But what a matter wave carry, at atomic level it is electron. So first it was thought that square of amplitude is current density. Later it was conceptualised that intensity is dependent on number of particles or current, so square of amplitude of matter wave is probability of particle. If it is operated with space, then its gives probability in terms of location or chances of finding at given region. And the amplitude became probability amplitude.


As we seen that intensity of light gives concept of square of amplitude as probability. Now question is probability of what, position, momentum, energy. As Born's interpretation was about wave mechanics of Schrodinger and not about matrix mechanics of Heisenberg. It says that $\psi$ itself has no meaning but $|\psi|^2$ because real part of complex function gives magnitude or length, a scalar that is needed at the end of day.

The crucial point is that in wave mechanics, state is stationary and operator is dynamic that lead edge to matrix mechanics which is opposite of it. This means that there is once amplitude or probability needed to draw for given wave function and then it gives average value of other observables operated on wave function. In my view, this dynamic operator and stationary state is bit loose because average value is given by operator operate on state, changing it state and then original state is correlated with changed state gives average value.


In the Born interpretation, $\psi$ doesn't have any physical meaning. It's only $|\psi|^2$ that has a physical meaning. In the case of a one dimensional wave function, $|\psi(x)|^2$ is a probability density where the probability of finding the particle described by that wave function between two points $x=a$ and $x=b$ is $$P(a \leq x \leq b)=\int_a^b|\psi|^2dx$$

provided the wave function has been normalized so that $$\int_{-\infty}^{\infty}|\psi|^2dx=1$$

We don't care about $|\psi|$ in the Born interpretation even if it is real and non-negative.

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    $\begingroup$ Thank you @Muster_Mark for the reply. But this still doesnt answer as to why square of modulus was chosen. Does not modulus of wave function have a physical significance ? $\endgroup$ Commented Jul 20, 2015 at 14:25

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