A.30 The evolution of probability

This note looks at conservation of probability, and the resulting definitions of the reflection and transmission coefficients in scattering. It also explains the concept of the “probability current” that you may occasionally run into.

For the unsteady Schrödinger equation to provide a physically correct description of nonrelativistic quantum mechanics, particles should not be able to disappear into thin air. In particular, during the evolution of the wave function of a single particle, the total probability of finding the particle if you look everywhere should stay one at all times:

$$\int_{x=-\infty}^{\infty} \vert\Psi\vert^2 {\,\rm d}x = 1 \mbox{ at all times}$$

Fortunately, the Schrödinger equation

$${\rm i}\hbar \frac{\partial\Psi}{\partial t} = - \frac{\hbar^2}{2m} \frac{\partial^2\Psi}{\partial x^2} + V \Psi$$

does indeed conserve this total probability, so all is well.

To verify this, note first that $\vert\Psi\vert^2=\Psi^*\Psi$, where the star indicates the complex conjugate, so

$$\frac{\partial\vert\Psi\vert^2}{\partial t} = \Psi^*\fra... ...l\Psi}{\partial t} + \Psi\frac{\partial\Psi^*}{\partial t}$$

To get an expression for that, take the Schrödinger equation above times $\Psi^*/{\rm i}\hbar$ and add the complex conjugate of the Schrödinger equation,

$$- {\rm i}\hbar \frac{\partial\Psi^*}{\partial t} = - \fr... ...hbar^2}{2m} \frac{\partial^2\Psi^*}{\partial x^2} + V \Psi^*,$$

times $-\Psi/{\rm i}\hbar$. The potential energy terms drop out, and what is left is

$$\frac{\partial\vert\Psi\vert^2}{\partial t} = \frac{{\rm... ...x^2} - \Psi\frac{\partial^2\Psi^*}{\partial x^2} \right).$$

Now it can be verified by differentiating out that the right hand side can be rewritten as a derivative:

$$\frac{\partial\vert\Psi\vert^2}{\partial t} = - \frac{\par... ...ial x} - \Psi^*\frac{\partial\Psi}{\partial x} \right) %$$ (A.231)

For reasons that will become evident below, $J$ is called the “probability current.” Note that $J$, like $\Psi$, will be zero at infinite $x$ for proper, normalized wave functions.

If (A.231) is integrated over all $x$, the desired result is obtained:

$$\frac{{\rm d}}{{\rm d}t} \int_{x=-\infty}^{\infty}\vert\Psi\vert^2 {\,\rm d}x = - J\Big\vert _{x=-\infty}^{\infty} = 0.$$

Therefore, the total probability of finding the particle does not change with time. If a proper initial condition is provided to the Schrödinger equation in which the total probability of finding the particle is one, then it stays one for all time.

It gets a little more interesting to see what happens to the probability of finding the particle in some given finite region $a\le{x}\le{b}$. That probability is given by

$$\int_{x=a}^b \vert\Psi\vert^2 {\,\rm d}x$$

and it can change with time. A wave packet might enter or leave the region. In particular, integration of (A.231) gives

$$\frac{{\rm d}}{{\rm d}t} \int_{x=a}^b\vert\Psi\vert^2 {\,\rm d}x = J_a - J_b$$

This can be understood as follows: $J_a$ is the probability flowing out of the region $x<a$ into the interval $[a,b]$ through the end $a$. That increases the probability within $[a,b]$. Similarly, $J_b$ is the probability flowing out of $[a,b]$ at $b$ into the region $x>b$; it decreases the probability within $[a,b]$. Now you see why $J$ is called probability current; it is equivalent to a stream of probability in the positive $x$-direction.

The probability current can be generalized to more dimensions using vector calculus:

$$\vec J = \frac{{\rm i}\hbar}{2m} \left(\Psi\nabla\Psi^* - \Psi^*\nabla\Psi \right) %$$ (A.232)

and the net probability flowing out of a region is given by

$$\int \vec J \cdot {\vec n}{\,\rm d}A %$$ (A.233)

where $A$ is the outside surface area of the region, and ${\vec n}$ is a unit vector normal to the surface. A surface integral like this can often be simplified using the divergence (Gauss or whatever) theorem of calculus.

Returning to the one-dimensional case, it is often desirable to relate conservation of probability to the energy eigenfunctions of the Hamiltonian,

$$- \frac{\hbar^2}{2m} \frac{{\rm d}^2\psi}{{\rm d}x^2} + V \psi = E \psi$$

because the energy eigenfunctions are generic, not specific to one particular example wave function $\Psi$.

To do so, first an important quantity called the “Wronskian” must be introduced. Consider any two eigenfunctions $\psi_1$ and $\psi_2$ of the Hamiltonian:

$$\begin{eqnarray*} & \displaystyle - \frac{\hbar^2}{2m} \frac{{\rm d}^2\psi_1... ...{2m} \frac{{\rm d}^2\psi_2}{{\rm d}x^2} + V \psi_2 = E \psi_2 \end{eqnarray*}$$

If you multiply the first equation above by $\psi_2$, the second by $\psi_1$ and then subtract the two, you get

$$\frac{\hbar^2}{2m} \left( \psi_1\frac{{\rm d}^2\psi_2}... ...2} - \psi_2\frac{{\rm d}^2\psi_1}{{\rm d}x^2} \right) = 0$$

The constant $\hbar^2/2m$ can be divided out, and by differentiation it can be verified that the remainder can be written as

$$\frac{{\rm d}W}{{\rm d}x} = 0 \qquad \mbox{where } W =... ...rm d}\psi_2}{{\rm d}x} - \psi_2\frac{{\rm d}\psi_1}{{\rm d}x}$$

The quantity $W$ is called the Wronskian. It is the same at all values of $x$.

As an application, consider the example potential of figure A.11 in addendum {A.25} that bounces a particle coming in from the far left back to where it came from. In the left region, the potential $V$ has a constant value $V_{\rm {l}}$. In this region, an energy eigenfunction is of the form

$$\psi_E = C^{\rm {l}}_{\rm {f}} e^{{\rm i}p_{\rm {c}}^{\rm ... ...d\mbox{where } p_{\rm {c}}^{\rm {l}}=\sqrt{2m(E-V_{\rm {l}})}$$

At the far right, the potential grows without bound and the eigenfunction becomes zero rapidly. To make use of the Wronskian, take the first solution $\psi_1$ to be $\psi_E$ itself, and $\psi_2$ to be its complex conjugate $\psi_E^*$. Since at the far right the eigenfunction becomes zero rapidly, the Wronskian is zero there. And since the Wronskian is constant, that means it must be zero everywhere. Next, if you plug the above expression for the eigenfunction in the left region into the definition of the Wronskian and clean up, you get

$$W = \frac{2{\rm i}p_{\rm {c}}^{\rm {l}}}{\hbar} \left(\v... ...}_{\rm {b}}\vert^2-\vert C^{\rm {l}}_{\rm {f}}\vert^2\right).$$

If that is zero, the magnitude of $C^{\rm {l}}_{\rm {b}}$ must be the same as that of $C^{\rm {l}}_{\rm {f}}$.

This can be understood as follows: if a wave packet is created from eigenfunctions with approximately the same energy, then the terms $C^{\rm {l}}_{\rm {f}}e^{{{\rm i}}p_{\rm {c}}^{\rm {l}}x/\hbar}$ combine for large negative times into a wave packet coming in from the far left. The probability of finding the particle in that wave packet is proportional to the integrated square magnitude of the wave function, hence proportional to the square magnitude of $C^{\rm {l}}_{\rm {f}}$. For large positive times, the $C^{\rm {l}}_{\rm {b}}e^{-{{\rm i}}p_{\rm {c}}^{\rm {l}}x/\hbar}$ terms combine in a similar wave packet, but one that returns towards the far left. The probability of finding the particle in that departing wave packet must still be the same as that for the incoming packet, so the square magnitude of $C^{\rm {l}}_{\rm {b}}$ must be the same as that of $C^{\rm {l}}_{\rm {f}}$.

Next consider a generic scattering potential like the one in figure 7.22. To the far left, the eigenfunction is again of the form

$$\psi_E = C^{\rm {l}}_{\rm {f}} e^{{\rm i}p_{\rm {c}}^{\rm ... ...d\mbox{where } p_{\rm {c}}^{\rm {l}}=\sqrt{2m(E-V_{\rm {l}})}$$

while at the far right it is now of the form

$$\psi_E = C^{\rm {r}} e^{{\rm i}p_{\rm {c}}^{\rm {r}}x/\hba... ...d\mbox{where } p_{\rm {c}}^{\rm {r}}=\sqrt{2m(E-V_{\rm {r}})}$$

The Wronskian can be found the same way as before:

$$W = \frac{2{\rm i}p_{\rm {c}}^{\rm {l}}}{\hbar} \left(\v... ...{\rm i}p_{\rm {c}}^{\rm {r}}}{\hbar} \vert C^{\rm {r}}\vert^2$$

The fraction of the incoming wave packet that ends up being reflected back towards the far left is called the “reflection coefficient” $R$. Following the same reasoning as above, it can be computed from the coefficients in the far left region of constant potential as:

$$R = \frac{\vert C^{\rm {l}}_{\rm {b}}\vert^2}{\vert C^{\rm {l}}_{\rm {f}}\vert^2}$$

The reflection coefficient gives the probability that the particle can be found to the left of the scattering region at large times.

Similarly, the fraction of the incoming wave packet that passes through the potential barrier towards the far right is called the “transmission coefficient” $T$. It gives the probability that the particle can be found to the right of the scattering region at large times. Because of conservation of probability, $T=1-R$.

Alternatively, because of the Wronskian expression above, the transmission coefficient can be explicitly computed from the coefficient of the eigenfunction in the far right region as

$$T = \frac{p_{\rm {c}}^{\rm {r}}\vert C^{\rm {r}}\vert^2} ... ...{l}})} \quad p_{\rm {c}}^{\rm {r}}=\sqrt{2m(E-V_{\rm {r}})}$$

If the potential energy is the same at the far right and far left, the two classical momenta are the same, $p_{\rm {c}}^{\rm {r}}=p_{\rm {c}}^{\rm {l}}$. Otherwise, the reason that the ratio of classical momenta appears in the transmission coefficient is because the classical momenta in a wave packet have a different spacing with respect to energy if the potential energy is different. (The above expression for the transmission coefficient can also be derived explicitly using the Parseval equality of Fourier analysis, instead of inferred from conservation of probability and the constant Wronskian.)