What I understood is that probability amplitude is the square root of the probability of finding an electron around a nucleus, but the square root of the probability does not mean anything in the physical sense.

Can any please explain the physical significance of the probability amplitude in quantum mechanics?

I read the Wikipedia article on probability amplitude many times over. What are those dumbbell shaped images representing?


6 Answers 6


Part of you problem is

"Probability amplitude is the square root of the probability [...]"

The amplitude is a complex number whose amplitude is the probability. That is $\psi^* \psi = P$ where the asterisk superscript means the complex conjugate.1 It may seem a little pedantic to make this distinction because so far the "complex phase" of the amplitudes has no effect on the observables at all: we could always rotate any given amplitude onto the positive real line and then "the square root" would be fine.

But we can't guarantee to be able to rotate more than one amplitude that way at the same time.

More over, there are two ways to combine amplitudes to find probabilities for observation of combined events.

  • When the final states are distinguishable you add probabilities: $P_{dis} = P_1 + P_2 = \psi_1^* \psi_1 + \psi_2^* \psi_2$.

  • When the final state are indistinguishable,2 you add amplitudes: $\Psi_{1,2} = \psi_1 + \psi_2$, and $P_{ind} = \Psi_{1,2}^*\Psi_{1,2} = \psi_1^*\psi_1 + \psi_1^*\psi_2 + \psi_2^*\psi_1 + \psi_2^*\psi_2$. The terms that mix the amplitudes labeled 1 and 2 are the "interference terms". The interference terms are why we can't ignore the complex nature of the amplitudes and they cause many kinds of quantum weirdness.

1 Here I'm using a notation reminiscent of a Schrödinger-like formulation, but that interpretation is not required. Just accept $\psi$ as a complex number representing the amplitude for some observation.

2 This is not precise, the states need to be "coherent", but you don't want to hear about that today.


Before trying to understand quantum mechanics proper, I think it's helpful to try to understand the general idea of its statistics and probability.

There are basically two kinds of mathematical systems that can yield a nontrivial formalism for probability. One is the kind we're familiar with from everyday life: each outcome has a probability, and those probabilities directly add up to 100%. A coin has two sides, each with 50% probability. $50\% + 50\% = 100\%$, so there you go.

But there's another system of probability, very different from what you and I are used to. It's a system where each event has an associated vector (or complex number), and the sum of the squared magnitudes of those vectors (complex numbers) is 1.

Quantum mechanics works according to this latter system, and for this reason, the complex numbers associated with events are what we often deal with. The wavefunction of a particle is just the distribution of these complex numbers over space. We have chosen to call these numbers the "probability amplitudes" merely as a matter of convenience.

The system of probability that QM follows is very different from what everyday experience would expect us to believe, and this has many mathematical consequences. It makes interference effects possible, for example, and such is only explainable directly with amplitudes. For this reason, amplitudes are physically significant--they are significant because the mathematical model for probability on the quantum scale is not what you and I are accustomed to.

Edit: regarding "just extra stuff under the hood." Here's a more concrete way of talking about the difference between classical and quantum probability.

Let $A$ and $B$ be mutually exclusive events. In classical probability, they would have associated probabilities $p_A$ and $p_B$, and the total probability of them occurring is obtained through addition, $p_{A \cup B} = p_A + p_B$.

In quantum probability, their amplitudes add instead. This is a key difference. There is a total amplitude $\psi_{A \cup B} = \psi_A + \psi_B$. and the squared magnitude of this amplitude--that is, the probability--is as follows:

$$p_{A \cup B} = |\psi_A + \psi_B|^2 = p_A + p_B + (\psi_A^* \psi_B + \psi_A \psi_B^*)$$

There is an extra term, yielding physically different behavior. This quantifies the effects of interference, and for the right choices of $\psi_A$ and $\psi_B$, you could end up with two events that have nonzero individual probabilities, but the probability of the union is zero! Or higher than the individual probabilities.

  • $\begingroup$ I'm not too happy with the formulation of "mathematical systems that can yield a nontrivial formalism for probability." Firstly, becuase it sounds like you imply that there are only these two "systems", and secondly, because the quantum framework is still one where "each outcome has a probability, and those probabilities directly add up to 100%." It's just extra dynamics under the hood. $\endgroup$
    – Nikolaj-K
    Mar 21, 2013 at 16:13
  • 3
    $\begingroup$ There are only these two systems. It is mathematically proven that you couldn't have, say, an amplitude that must be raised to the 4th power. There is only classical probability as we know it and the quantum kind. It's not just extra stuff under the hood, either. See my edit. $\endgroup$
    – Muphrid
    Mar 21, 2013 at 16:28
  • 3
    $\begingroup$ Whatever is mathematically proven must be w.r.t. some postulates and these are not stated. Also, there are the observable who's probabilities sum to 100% (namely the probability to be in any of a total set of eigenstates) and in this sense it's just probability theory with complex dynamics under the hood. I still don't think this is an inappropriate formulation. $\endgroup$
    – Nikolaj-K
    Mar 21, 2013 at 18:23
  • 1
    $\begingroup$ @Muphrid could you provide a reference for the result that "there are only those two systems"? $\endgroup$
    – glS
    Feb 3, 2015 at 21:54
  • 2
    $\begingroup$ "In quantum probability, their amplitudes add instead. This is a key difference." -- This doesn't make sense to me. Suppose $p_A = 1/\sqrt{2}$ and $p_B = 1/\sqrt{2}$. Then $|1/\sqrt{2}+1/\sqrt{2}|^2 = 2$. What does this mean? $\endgroup$ Mar 29, 2017 at 23:06

In quantum mechanics, the amplitude $\psi$, and not the probability $\mid\psi\mid^2$, is the quantity which admits the superposition principle. Notice that the dynamics of the physical system (Schrödinger equation) is formulated in terms of and is linear in the evolution of this object. Observe that working with superposition of $\psi$ also permits complex phases $e^{i\theta}$ to play a role. In the same spirit, the overlap of two systems is computed by investigation of the overlap of the amplitudes.

  • 4
    $\begingroup$ All you say is factually correct, but since the question asked for an explanation in layman's terms I think there needs to be more explanation. $\endgroup$
    – user9886
    Mar 21, 2013 at 16:21
  • 1
    $\begingroup$ @user9886: The integrals involving position operators are layman's terms? $\endgroup$
    – Nikolaj-K
    Mar 21, 2013 at 18:11
  • 1
    $\begingroup$ What is the benefit in using complex phases rather than just sine and cosine? $\endgroup$ Feb 27, 2014 at 2:57

I agree with the other answers provided. However, you may find the probability amplitudes more intuitive in the context of the Feynman path integral approach.

Suppose a particle is created at the location $x_1$ at time $0$ and that you want to know the probability for observing it later at some position $x_2$ at time $t$.

Every path $P$ that starts at $x_1$ at time zero and ends at $x_2$ at time $t$ is associated with a (complex) probability amplitude $A_P$. Within the path integral approach, the total amplitude for the process initially described is given by the sum of all these amplitudes:

$A_{\textrm{total}} = \sum_P A_P$

I.e. the sum over all possible paths the particle could take between $x_1$ and $x_2$. These paths interfere coherently, and the probability for observing the particle at $x_2$ at time $t$ is given by the square of the total amplitude:

$\textrm{probability to observe the particle at $x_2$ at time $t$} = |A_{\textrm{total}}|^2 = |\sum_P A_P|^2$

I should note that the Feynman path integral formalism (described above) is actually a special case of a more general approach wherein the amplitudes are associated with processes rather than paths.

Also, a good reference for this is volume 3 of The Feynman Lectures.


In quantum mechanics a particle is described by its wave-function $\psi$ (in spatial representation it would for example be $\psi(x,t)$, but I omit the arguments in the following). Observables, like the position $x$ are represented by operators $\hat x$. The mean value of the position of an particle is calculated as $$\int \mathrm{d}x \tilde \psi \hat x \psi.$$

Since $\hat x$ applied to $\psi(x,t)$ just gives the position $x$ times $\psi(x,t)$ we can write the integral as $$\int \mathrm{d}x x \tilde \psi \psi.$$

$\tilde \psi$ is the complex conjugate of $\psi$ and therefore $\tilde \psi \psi=|\psi|^2$.

And finally, since a mean value is usually computed as an integral over the variable times a probability distribution $\rho$ as $$\langle X \rangle_\rho=\int \mathrm{d}X X \rho(X)$$ $|\psi|^2$ can be interpreted as a probability density of finding the particle at some point. E.g. The probability of it being between $a$ and $b$ is $$\int_a^b\mathrm{d}x|\psi|^2$$

So the wave function (which is the solution to the Schrödinger equation that describes the system in question) is a probability amplitude in the sense of the first sentence of the article you linked.

Lastly, the dumbbell shows the area in space where $|\psi|^2$ is larger than some very small number, so basically the regions, where it is not unlikely to find the electron.


Have a look at this simplified statement in describing the behavior of a particle in a potential problem:

In quantum mechanics, a probability amplitude is a complex number whose modulus squared represents a probability or probability density.

This complex number comes from a solution of a quantum mechanical equation with the boundary conditions of the problem, usually a Schroedinger equation, whose solutions are the "wavefunctions" $\psi(x)$, where $x$ represents the coordinates generically for this argument.

The values taken by a normalized wave function $\psi$ at each point $x$ are probability amplitudes, since $|\psi(x)|^2$ gives the probability density at position $x$.

To get from the complex numbers to a probability distribution, the probability of finding the particle, we have to take the complex square of the wavefunction $\psi^\ast\psi$ .

So the "probability amplitude" is an alternate definition/identification of "wavefunction", coming after the fact, when it was found experimentally that $\psi^\ast \psi$ gives a probability density distribution for the particle in question.

First one computes ψ and then one can evaluate the probability density $\psi^\ast \psi$, not the other way around. The significance of $\psi$ is that it is the result of a computation.

I agree it is confusing for non physicists who know probabilities from statistics.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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