I asked a related question to this here:

Why are transition amplitudes more fundamental than probabilities in quantum mechanics?

It was closed as a duplicate, but unfortunately the answers in related links didn't help me answer my question. I will try again and hope to do better.

In classical mechanics, if I have an object, it has a definite position $x$. We don't have to guess, we know where it is. But quantum mechanics doesn't seem to work that way.

Consider a particle with a probability density $\rho (x,t)$ for some given $(x,t)$. As I understand it, $\rho$ represents the probability of finding a particle at position $x$ given a measurement at time $t$. We still might not find it! But if we perform this procedure on an ensemble of identically prepared systems, then we should on average expect to find the particle at $x$. I hope I am understanding this correctly.

Now, consider a standard die or coin. There are a set of possible outcomes. For instance $\lbrace H,T\rbrace$ for a coin. Each possible outcome has a probability. And these probabilities are fixed. So the analog for a die's probability space is the position space for a particle of possible positions at a time $t$. But does the physical reality of a particle's probability density only have meaning at the point of measurement?

In other words, consider now a set of $n$ 6-side dice. We will imagine that as I move through time, my die $i$ may morph into a die $j$ and have a different probability distribution over the same set of outcomes. In other words, we have differently bias dice, differently "loaded". Suppose that at time $t_i$ if I roll it, I have a the probabilities given die $i$. But if I roll at some later time $t_j$, then I will have different die $j$ and thus a different set of probabilities. Same set of outcomes, but different probabilities. One could imagine that a particle is like this. If I measure it at time $t_i$ I get a probability distribution $\rho_i$ but if I measure at a later time $t_j$, then it could be a different distribution $\rho_j$. But the set of possible $\rho$ is known. So if we had a function $f$ that would tell us given $t_i < t_j$ and given $\rho_i$, then $f(\rho_i)=\rho_j$, then the probability densities would have meaning beyond the single moment $t_i$. Then they would give us information about future probability densities. So although we never know where the particle is, we know where it might be. Is this how I should interpret the probability densities? Or are we saying no such $f$ exists. That is, are we saying that given $(x,t_i)$ and $\rho(x,t_i)$, there is no way of knowing what the probability density is at a later time $t_j$. If that's what we are saying, then here's my question:

My Question:

If we cannot infer a later probability density $\rho_j$ given a current one $\rho_i$, how do we describe the time evolution of the system in a way that gives us information about where the particle might be in the future?

  • $\begingroup$ Do the dice in your example obey the Schroedinger equation? It is this equation that allows us to do virtually all calculations in quantum mechanics. $\endgroup$
    – Ryan Unger
    Feb 14, 2015 at 21:35
  • $\begingroup$ I think this is maybe what I am not understanding. I don't understand what the Schrödinger equation tells us about the ability to measure current and future probability densities. As you can see, I haven't mentioned probability amplitudes anywhere here. So I clearly don't fully understand why they are important. Lol I didn't try to for instance draw an analogy for a die probability amplitude $\endgroup$ Feb 14, 2015 at 21:54
  • $\begingroup$ I'm not comfortable answering yet. Let me propose something simpler. Say we have two coins -- one perfectly even and one that gets heads 75% of the time. Suppose that the first coin turns into the second one at some time $t$. So you want to know if, given the probability distribution of the bad coin at time $>t$, can we infer that it was a good coin at time $<t$? $\endgroup$
    – Ryan Unger
    Feb 14, 2015 at 22:39
  • $\begingroup$ Yes. That's the idea. And if the reverse is also possible. Given the good distribution, can I predict that the bad distribution is next? If you would like to propose an edit, I would accept it. The question as stated may be worded in a way that will reduce the value for future readers. $\endgroup$ Feb 14, 2015 at 22:50
  • $\begingroup$ Tangential to your question, but "But does the physical reality of a particle's probability density only have meaning at the point of measurement?" has two misconceptions worth to point out: probability density and $\psi$ have no physical reality. They are mathematical functions assigned and operated by humans and they can change when the person using it decides to do so. $\endgroup$ Feb 15, 2015 at 13:52

2 Answers 2


For your dice example in the question or my coin example in the comments, the answer is that we cannot predict the probability distribution in the future. I think this is intuitively clear. Say we have some sketchy magician. He flips a coin a bunch of times and we get a roughly 50/50 split of heads and tails. His partner distracts us and while we look away he switches the coin to one that is "loaded," i.e. does not have a 1:1 heads to tails distribution. There is no reason for us to believe anything has changed. Even if he tells us that he switched the coins, there is absolutely no way for us to know what the new distribution is.

You ask, "So what about particles and quantum mechanics?" This is an entirely different matter. Quantum mechanical probabilities are understood in terms of a probability amplitude $\psi(\mathbf{x},t)$ with the understanding that the probability distribution is given by $\rho(\mathbf{x},t)=\psi^*(\mathbf{x},t)\psi(\mathbf{x},t)$. The key feature of the probability amplitude is that it is obeys the Schroedinger equation $$i\frac{\partial\psi}{\partial t}=H\psi$$ This equation determines the evolution of the system.

Let us examine why the Schroedinger equation is important for PDFs.

1: The wave function is an energy eigenstate. This means that $\psi$ obeys the time-independent Schroedinger equation$^1$ $$H\psi=E\psi$$ From the time-dependent equation we get $$i\frac{\partial\psi}{\partial t}=E\psi$$ Hence, the time dependence of a Hamiltonian eigenstate is given by$^2$ $$\psi(\mathbf{x},t)=\phi(\mathbf{x})e^{-iEt}$$ Thus, perhaps surprisingly, the probability density of an eigenstate is constant in time.

2: The wave function is not an energy eigenstate. For such a wave function the time dependence is more complicated: We must determine the propagator. The time evolution of the probability density is given by the continuity equation $$\frac{\partial\rho}{\partial t}=-\frac{1}{m}\nabla\cdot\mathfrak{I}(\psi^*\nabla\psi)$$ which is itself derived from the Schroedinger equation$^3$.

We see thus that our knowledge of the probability density stems from the Schroedinger equation. Without such an overall governing principle, there is no way to predict future probability densities.

$^1$ In my opinion, calling this mere eigenvalue equation a Schroedinger equation is a bit much.

$^2$ The first factor on the right is often interpreted as the wave function of the initial data.

$^3$ See, e.g., Ballentine, Quantum Mechanics: A Modern Development (1998), sect. 4.4.

  • 2
    $\begingroup$ Nice answer. Additionally (because OP seems confused about why we consider $\psi$ at all if it is $\rho$ we are after) it has to be said that $\rho$, as a real function, has less degress of freedom than $\psi$, and therefore it does not suffice to look at $\rho$ alone. In particular, there is no derivable evolution equation that could be written in terms of $\rho$ only. $\endgroup$
    – ACuriousMind
    Feb 15, 2015 at 0:17
  • 1
    $\begingroup$ @ACuriousMind: I completely forgot to mention that. In particular, since we lose information by taking the square root of the PDF ($\sqrt{|\psi|^2}\ne\psi$), we cannot recover the phase of the wave function. Since the probability currents depends on the imaginary part of $\psi$, we need to write the evolution equation in terms of $\psi$, not $\rho$. $\endgroup$
    – Ryan Unger
    Feb 15, 2015 at 0:26
  • $\begingroup$ What degrees of freedom does $\psi$ have that $\rho$ doesn't? $\endgroup$ Feb 15, 2015 at 0:40
  • 1
    $\begingroup$ @StanShunpike: Phase. $\psi=\sqrt{\rho}e^{i\theta}$. $\endgroup$
    – Ryan Unger
    Feb 15, 2015 at 0:47
  • $\begingroup$ What does $\mathfrak{I}$ stand for? Thanks again btw for your posts. They're great. $\endgroup$ Feb 15, 2015 at 2:15

The short answer is that we can't. You said it yourself!

"If we cannot infer a later probability density $\rho_{j}$ given a current one $\rho_{i}$, how do we describe the time evolution of the system..."

If I am to be within the premise of the formulation we do not have knowledge of some biased probability distribution at a later time. At best we have the uniform probability distribution, such that our model would be that the particle is equally likely to be anywhere (in some given volume). If we don't know things exactly we resort to probability distributions and if we do not have some information on which we can construct a biased probability distribution, our power of prediction is non-existent.


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