# What is the difference in Probability Density and probability of a particle being in a particular state?

I'm just starting to learn Quantum Mechanics, and this question is confusing me: we say that the probability of finding a particle in a state given by the Eigenstate $$o_i$$ is the modulus squared of the eigenvalue of the eigenstate, call it $$|c_n|^2$$. What is the difference in $$|c_n|^2$$ and the probability density $$\rho$$? What is the physical interpretation of $$\rho$$?

• If your question is just about the difference between probability and probability density, it may be a better fit on Mathematics. – Sandejo Oct 5 '20 at 2:00
• It's about the difference in probability of a certain state and probability density, as well as the physical interpretation of probability density. – Daniel Waters Oct 5 '20 at 2:18

The difference has to do with the nature of the particular observable. Imagine for example a photon whose polarization we are interested in measuring. The photon's polarization can be in either of two states, say horizontal $$|H\rangle$$ or vertical $$|V\rangle$$ (of course, you are free to choose a different basis). Then we can describe the photon's state as a superposition of these states: $$|\psi\rangle = \alpha|H\rangle +\beta|V\rangle$$ Where $$\alpha$$ and $$\beta$$ are the probability amplitudes for states $$|H\rangle$$ and $$|V\rangle$$ respectively. The probability of finding the photon in one of these states is then $$\alpha^2$$ and $$\beta^2$$, respectively. These values are constrained by the relation $$\alpha^2+\beta^2 =1$$, because the photon must be in one of these two polarization states.

Okay, I suspect that this all seems well and good, but how does it relate to the probability density? Well, in the above scenario, our observable had only a discrete number of possible values: $$|H\rangle$$ or $$|V\rangle$$. While this example uses two, it could in theory be any number. In the general case, we could say: $$|\psi\rangle = \sum_ic_i|\Psi_i\rangle$$ Where the probabilities amplitudes $$c_i\in\mathbb{C}$$ are constrained such that $$\sum_i|c_i|^2=1$$.

Not all observables can be discretized in this way, though; for example, consider a particle's position. How can we assign a probability to each potential position a particle could have? After all, a continuous variable can have any of an infinite number of values, and so we would expect the probability of finding a particle at any exact location to be zero.

To solve this problem, we instead talk about the probability of finding the particle not at some position, but rather in some interval. So long as the interval has nonzero length, it is reasonable to expect the probability of finding the particle within that interval to also be nonzero.

Consider a small interval $$[x,x+\mathrm{d}x]$$ of width $$\mathrm{d}x$$. We can associate with this small interval a small probability $$\mathrm{d}P$$. In the limit as $$\mathrm{d}x\to0$$, one finds that $$\mathrm{d}P\propto\mathrm{d}x$$. The associated proportionality constant, let us call it $$\rho$$, is itself a function of position: $$\mathrm{d}P=\rho(x)~\mathrm{d}x$$.

Our proportionality constant $$\rho$$ is the probability density, and it tells us how much probability to associate with a given interval. You can make a physical analogy to any other type of density. For example, the linear mass density of a string would tell you how much of the string's mass is associated with any given segment of its length. It has dimensions of inverse length (probability per length, but probability is dimensionless).

One can see how we can integrate our above equation over some interval $$[a,b]$$ to find the associated probability: $$P(a,b)=\int_a^b\mathrm{d}x~\rho(x)$$ And as I suspect you know, this probability density is the square of the modulus of the wavefunction, giving us: $$P(a,b)=\int_a^b\mathrm{d}x~\psi^*\psi$$

I myself am currently a student of quantum mechanics, but as far as I can see, one can consider the probability density to be the continuous limit of the squares of the moduli of the probability amplitudes in a close packing of discrete states (i.e. $$|\psi\rangle = \lim_{(x_{i+i}-x_i)\to0}\sum_i^\infty c_i|x_i\rangle$$ where $$|x_i\rangle$$ is a pure state of position corresponding to the particle being located at $$x_i$$).

You use probability for a finite number of states. E.G. An electron has a 50% probability of being spin up.

For an infinite number of states, you need probability density. Suppose a particle is confined to a 1D potential well from -1 to 1. The probability of being at any particular x is 0. You can't learn much from that probability, so you use density.

If you divide the range into n pieces of length L, and the probability is of being in any piece is $$p = 1/n = L/2$$, the probability distribution is uniform. As $$n\to \infty$$, the probability of being in any piece $$p \to 0$$. But probability/length stays fixed. $$p/L = (1/n)/L =1/2$$. This is the probability density, $$\rho$$, the probability per unit length.

The important property: The probability of being in a piece of length L is $$p = \rho L$$ for a uniform density. In general $$p = \int \rho \space dl$$.