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First, let's say I have a classical system involving throwing a fair coin. There are two possible events $\{\text{head},\text{tails}\}$. Their respective probabilities are:

$$ P(\text{head})=\frac{1}{2}\\ P(\text{tails})=\frac{1}{2}\\ P(\text{head})+P(\text{tails})=1 $$

In a quantum system scenario, the probabilities are replaced by a complex amplitude. The square modulus of the amplitude is a "probability density" evaluated at a point in phase space. For instance:

$$ A(\text{head})=c_1e^{i\theta_1}\\ A(\text{tails})=c_2e^{i\theta_2}\\ A(\text{head})+A(\text{tails})=c_1e^{i\theta_1}+c_2e^{i\theta_2}\\ I=(c_1e^{i\theta_1}+c_2e^{i\theta_2})(c_1e^{-i\theta_1}+c_2e^{-i\theta_2})=c_1^2+c_2^2 + 2c_1c_2 \cos (\theta_2-\theta_2) $$

And the sum of probabilities is given by the integral over all of phase space:

$$ \int_{-\infty}^\infty I(c_1[x],c_2[x],\theta_1[x],\theta_2[x])dx=1 $$

where $x$ is a prametrization for $c_1,c_2,\theta_1,\theta_2$.

Thus the quantum analog of

$$ P(\text{head})+P(\text{tails})=1 $$

is

$$ \int_{-\infty}^\infty I(c_1[x],c_2[x],\theta_1[x],\theta_2[x])dx=1 $$

Now, say I want to show that classical probability is a special case of quantum probability. I can set, as a restriction, the complex amplitude to be real, then the usual sum of probabilities is obtained without interference:

$$ |\Re[A_1]+\Re[A_2]|^2=|A_1|^2+|A_2|^2+2|A_1||A_2| $$

But I do not understand the procedure to get rid of the integral?

$$ \int_{-\infty}^{\infty}( |A_1|^2+|A_2|^2+2|A_1||A_2|) dx=1 $$

It seems that, at best, the integral may reduce to a continuous probability distribution in the classical case, such as a gaussian integral - is that the case?

Qmechanic
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Anon21
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    Possible duplicate: https://physics.stackexchange.com/q/116595/2451 – Qmechanic Jul 17 '20 at 15:06
  • “ the probabilities are replaced by a complex amplitude”. I don’t get it. No: probabilities are not replaced by complex numbers, no more than intensities are replaced by fields in classical theories. Probabilities are as different from complex numbers as intensities are different from fields. – ZeroTheHero Sep 01 '20 at 08:00

1 Answers1

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Quantum mechanics relies on the usual probability theory. Specifically, the modulus square of a wave function is conventional probability density: $$w(x) = \psi^*(x)\psi(x).$$ The difference between quantum theory and the probabilistic description of classical systems is that the latter operates with the probability density (e.g., via Fokker-Plank equation or rate equations), whereas the former formulates equations for wave function or density matrix (which is a generalization of the wave function).

To summarize: quantum probability theory is not a spacial case, but the same probability theory.

Update
Note that the classical random processes mentioned above are not the classical limit of quantum fluctuations. Indeed, the classical limit of the quantum probability is deterministic classical mechanics. When processes in classical physics are random it is for different reasons - e.g., Brownian motion of a particle interacting with an environment. In the quantum domain one then has two sources of randomness - due to the quantum nature of particles and due to the other random effects, like collisions with other particles, etc. In fact, this is precisely the case where the wave function is not sufficient and one has to use the density matrix. Such processes are described using quantum kinetic equations or *master equation.

Roger V.
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  • So the classical limit of quantum probabilities is real probability density function (and not individual events such as coin tosses)? – Anon21 Jul 17 '20 at 15:01
  • @AlexandreH.Tremblay Not quite - we are talking about different sources of randomness here (see the update in my answer). But mathematically it is the same probability theory. – Roger V. Jul 17 '20 at 15:22
  • But I do not understand how the square modulus of a single event gives a probability, yet the square modulus of a sum of two events gives a probability density. How can a simple sum convert a probability to a probability density? – Anon21 Jul 17 '20 at 15:24
  • It is either discrete probability or probability density in both cases. Probability density is what one uses when the probability distribution is continuous, and speaking about probability of a specific value doesn't make sense - one can only speak about probability of getting a value in a specified interval. It is really not about QM, but about the basic probability theory. – Roger V. Jul 17 '20 at 15:26
  • Okay, but If I have an event with complex amplitude $z_1e^{i\theta_1}$, then its square module is a probability: $|z_1e^{i\theta_1}|^2=|z_1|^2$. But, if I add two events its square module is now a probability density : $|z_1e^{i\theta_1}+z_2e^{i\theta_2}|^2=|z_1|^2+|z_2|^2+2|z_1||z_2|\cos(\theta_2-\theta_1)$. How can a simple addition causes a switcheroo in the definitions from probability to probability density? – Anon21 Jul 17 '20 at 15:30
  • Why do you think it is a density? It is still a probability. What is new is that in a classical system you would add probabilities: $w=w_1 + w_2$, whereas in a quantum case you add probability amplitudes, which results in appearance of the interference term: $w = w_1 + w_2 + 2|z_1||z_2|\cos(\theta_2-\theta_1$). – Roger V. Jul 17 '20 at 15:42
  • The probability at any phase must be zero because the integral of 1+2+2|1||2|cos(2−1) over all possible phases must be equal to 1, and phases are continuous. Consequently, I would think of as a probability density over the phase 2−1? – Anon21 Jul 17 '20 at 15:49
  • Also If I have a classical system I can add the probability $p_1+p_2=1$ and still produce a probability for each event individually as $p_1/(p_1+p_2)$. Whereas in the quantum case, the event $p_1^2/(p_1^2+p_2^2+2p_1p_2\cos(\theta_2-\theta_1))$ can exceed 1 for some value of $\theta_2-\theta_1$. So I can't even think of this as a probability. I am told to "project" the probability to a photon collector plate instead of thinking about it in terms of the two initial discreet events; but I do not understand, mathematically or even intuitively, how the continuum emerges from the interference term? – Anon21 Jul 17 '20 at 15:54