PHYS 385 Lecture 6 - Probability Current 6 - 1

PHYS 385 Lecture 6 - Probability current 6 - 1

Lecture 6 - Probability current

What's important: · continuity equation · probability current Text: Gasiorowicz, Chap. 3

We have now provided the motivation for the free-particle Schrödinger equation in one dimension 2 2 (x,t ) h (x,t) ih = - 2 . (one dimension) t 2m x The interpretation of the wavefunction is that its complex square is equal to the probability density at x and t: [probability density at x, t] = | (x,t)|2

Note that (x,t) has units (!) which depend on the dimensionality of the equation: in one dimension, (x,t) has units of [length]-1/2 in three dimensions, (x,t) has units of [length]-3/2.

Now, we can't just say "gee, it looks like a probability" - there are some properties which must be established. First, we turn to fluids (no longer part of most physics curricula) and derive the continuity equation.

Fluids

What is of interested to us here is how the local density of a fluid changes as it moves.

Let's consider a tube with cross sectional areas AL and AR at each end. The fluid enters at the left with speed vL and leaves at the right with speed vR.

The density need not be the same at each end, and we call 1 and 2 the number density (number per unit volume) of the particles in the fluid. More frequently in fluid mechanics courses, is the mass density.

If you completed a biologically-oriented freshman course in physics (see Lec. 28 in my PHYS101 lectures), you learned that the relation

1 A1 v1 = 2 A2 v2 holds when the number of particles (atoms, molecules, cars, whatever...) in the fluid is conserved. Let's write more general equation which allows for particles to come and go. We'll simplify the geometry of the fluid to be a cylinder:

At the left hand side, the number of particles entering the tube per unit area per unit time is called the flux f, which equals the density times the velocity v of the particles: f = v.

This identification is obviously OK on dimensional grounds, and you can confirm it by considering the motion of a group of particles starting at the left hand edge at the same time. The reasoning is: 1. Particles sweep out a volume Av Dt in time Dt. 2. The number of particles contained in this volume is Av Dt. 3. The number passing through the boundary per unit area per unit time must be Av Dt divided by ADt, or f = v.

Now, the flux may be different at each end, if the number of particles not conserved during transit, i.e. if particles break up or condense. Then, the difference in Av Dt between the two ends is the change in the number of particles DN:

-DN = RAvR Dt - LAvL Dt or

-DN = ( RvR - LvL) A Dt.

Note that a minus sign is required, as the number of particles inside the tube goes down as the flux goes up along the tube.

Divide the expression by A DL where DL is the length of the tube, to obtain

-DN / (A DL) = ( RvR - LvL) Dt / DL.

Now, to clean up. The density is = N / (A DL), so the left hand side is the change in density. We replace the product by the flux f, so that -D = D f • Dt / DL.

Transpose, and take the infinitesimal limit: ¶ /¶t = -¶f / ¶x. Equation of continuity (1)

The probability of finding a fluid particle in a given region is proportional to the density of particles in that region, so there should be a probability current corresponding to the fluid equation of continuity. What is the current?

Probability current

We have proposed that | (x,t)|2 is a probability density P(x,t) = | (x,t)|2, (2) much like the fluid number density. The time derivative of this probability is then:

¶P /¶t = ¶( * ) /¶t = (¶ * /¶t) + * (¶ /¶t), (3) where we have dropped the arguments from the wavefunction . The one-dimensional Schrödinger equation provides what we need for the second term on the RHS, and its complex conjugate 2 2 *( x,t) h *(x,t ) - ih = - 2 t 2m x provides the first term. Substituting, 2 2 2 2 ¶ 1 æç h ¶ * h ¶ ö P = 2 - * 2 ¶t ih è 2m ¶x 2m ¶x ø 1 2 ¶ æ ¶ ¶ * = - h * - ö i 2m ¶x è ¶x ¶x ø h Note the choice of minus signs. This has the same functional form as the equation of continuity, ¶P /¶t = - ¶j /¶x (4) it's just a question of defining the probability current j (same as the flux f, but for probability). It should be clear from the derivative on the RHS that the correspondence is æ ¶ ¶ *ö j = h * - . 2im è ¶x ¶x ø

For future reference, note that we can add the product V(x) (x,t) to the right hand side of the Schrödinger equation without harming the definition of the probability current.

Expectation values

So, we have a well-defined probability density and its current. We can now use P(x,t) to determine the expectation of a function f(x) just as we do for any probability distribution. Denoting the expectation value of a function f(x) as , we have = ò f(x) P(x,t) dx in one dimension. We'll generalize this momentarily.