16 The Continuity Equation

16.1 The Continuity Equation

Our starting point is the probability density $$\begin{array}{}\text{(16.3)}& P(x,t):=|\psi (x,t){|}^2.\end{array}$$ We would like to understand how the probability density depends on time $t$. Let us therefore first recall Schrödinger’s equation for the wave function and its complex conjugate, $$\begin{array}{}\text{(16.4)}& \begin{array}{rl}{\partial }_t\psi (x,t)& =-\frac{i}{\hslash }(-\frac{{\hslash }^2}{2m}{\partial }_x^2+V(x))\psi (x,t),\\ {\partial }_t\bar{\psi }(x,t)& =\frac{i}{\hslash }(-\frac{{\hslash }^2}{2m}{\partial }_x^2+V(x))\bar{\psi }(x,t),\end{array}\end{array}$$ where we recall that the potential $V(x)$ is a real analytic function. We can now compute the time derivative of the probability density, $$\begin{array}{}\text{(16.5)}& \begin{array}{rl}{\partial }_{t}P& ={\partial }_t|\psi {|}^2\\ & =\psi ({\partial }_t\bar{\psi })+\bar{\psi }({\partial }_t\psi )\\ & =\frac{i}{\hslash }\psi (-\frac{{\hslash }^2}{2m}{\partial }_x^2+V)\bar{\psi }-\frac{i}{\hslash }\bar{\psi }(-\frac{{\hslash }^2}{2m}{\partial }_x^2+V)\psi .\end{array}\end{array}$$ Note that the dependence on the potential $V(x)$ cancels out between the two terms and the remainder becomes $$\begin{array}{}\text{(16.6)}& \begin{array}{rl}{\partial }_{t}P& =\frac{\hslash }{2mi}(\psi {\partial }_x^2\bar{\psi }-\overline{\psi }{\partial }_x^2\psi )\\ & =\frac{\hslash }{2mi}{\partial }_x(\psi {\partial }_x\bar{\psi }-\bar{\psi }{\partial }_x\psi )\\ & =-{\partial }_{x}J,\end{array}\end{array}$$ where the symbol $J$, $$\begin{array}{}\text{(16.7)}& J:=\frac{\hslash }{2mi}(\overline{\psi }{\partial }_x\psi -\psi {\partial }_x\overline{\psi })\end{array}$$ is known as the “probability current density”. The result $$\begin{array}{}\text{(16.8)}& {\partial }_{t}P+{\partial }_{x}J=0,\end{array}$$ is known as the “continuity equation”.

16.2 Physical Interpretation

To understand the physical interpretation of $J(x,t)$ and the continuity equation it is convenient to integrate it over an interval. Let us define $$\begin{array}{}\text{(16.9)}& P_{ab}(t)={\int }_a^{b}P(x,t)\mathrm{d}x\end{array}$$ to be the probability to find the particle in the interval $a<x<b$. The time derivative of this probability can be expressed in terms of the probability current at the boundaries of the interval, $$\begin{array}{}\text{(16.10)}& \begin{array}{rl}\frac{d}{dt}{P}_{ab}(t)& ={\int }_a^b{\partial }_{t}P(t,x)\mathrm{d}x\\ & =-{\int }_a^b{\partial }_{x}J(t,x)\mathrm{d}x\\ & =J(a,t)-J(b,t).\end{array}\end{array}$$

This has the following interpretation:

• $J(x,t)$ is the rate that probability is “flowing” from left to right at $x$.

• The rate of change of the probability $P_{ab}(t)$ to find the particle in the interval $a<x<b$ is equal to the rate that probability is flowing in at the boundaries $x=a$ and $x=b$.

Put simply, this expresses the conservation of probability: probability cannot be created or destroyed but flows from one region to the next.

It is illuminating to consider the following limits:

1. Sending $a\to -\mathrm{\infty }$ and $b\to \mathrm{\infty }$, the equation becomes $$\begin{array}{}\text{(16.11)}& \frac{d}{dt}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}P(x,t)\mathrm{d}x=0,\end{array}$$ since if the wave function is normalizable then $\psi (x,t)\to 0$ and hence $J(x,t)\to 0$ as $x\to \pm \mathrm{\infty }$. We therefore recover the conservation of the total probability to find the probability. In particular, if the wave function is normalised at $t=t_0$, $$\begin{array}{}\text{(16.12)}& {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}P(x,t=t_0)\mathrm{d}x=1,\end{array}$$ this will remain normalised for $t>0$. This is an important property that you proved in one of the problems in the chapter where we first introduced the Schrödinger’s equation.

2. Setting $a=x$ and $b=x+\mathrm{d}x$ we find $$\begin{array}{}\text{(16.13)}& {\partial }_{t}P(x,t)\mathrm{d}x=J(x,t)-J(x+\mathrm{d}x,t).\end{array}$$ In the limit $\mathrm{d}x\to 0$, we recover the continuity equation ${\partial }_{t}P(x,t)+{\partial }_{x}J(x,t)$, which therefore expresses the local conservation of probability in the neighbourhood of the point $x$.

16.3 Example: Stationary Wave functions

Let us consider a stationary wave function, $$\begin{array}{}\text{(16.14)}& \psi (x,t)=\varphi (x)e^{-iEt/\hslash },\hat{H}\varphi (x)=E\varphi (x).\end{array}$$ The probability density and current are independent of time, $$\begin{array}{}\text{(16.15)}& \begin{array}{rl}P(x,t)& =|\varphi (x){|}^2,\\ J(x,t)& =\frac{\hslash }{2mi}(\overline{\varphi }(x){\partial }_x\varphi (x)-\varphi (x){\partial }_x\overline{\varphi }(x)).\end{array}\end{array}$$ The continuity equation tells us that ${\partial }_{x}J(x,t)=0$ and therefore $J(x,t)=J_0$ is constant. Furthermore, if $\varphi (x)$ is square-normalizable then $J(x,t)\to 0$ as $x\to \pm \mathrm{\infty }$. However, if $J(x,t)$ is constant, then it must vanish $J(x,t)=0$.

As an explicit example, suppose we have a Hamiltonian eigenfunction in the infinite potential well $0<x<L$, $$\begin{array}{}\text{(16.16)}& \varphi (x)=\sqrt{\frac{2}{L}}\sin \left(\frac{n\pi x}{L}\right),\end{array}$$ for some $n\in {\mathbb{Z}}_{>0}$. Since $\varphi (x)$ is a real function, $$\begin{array}{}\text{(16.17)}& \begin{array}{rl}J& =\frac{\hslash }{2mi}(\overline{\psi }{\partial }_x\psi -\psi {\partial }_x\overline{\psi })=\frac{\hslash }{2mi}(\varphi {\partial }_x\varphi -\varphi {\partial }_x\varphi )=0,\end{array}\end{array}$$ as required.

In fact, square-normalizable Hamiltonian eigenfunctions can always be chosen real (up to a constant phase) and this argument is completely general. It provides another proof of $J(x,t)=0$ for a square-normalisable stationary solution.

16.4 Example: Sum of Two Stationary Wave functions

Now consider the normalized sum of two stationary wave functions, $$\begin{array}{}\text{(16.18)}& \psi (x,t)=\frac{1}{\sqrt{2}}({\varphi }_1(x)e^{-i{E}_{1}t/\hslash }+{\varphi }_2(x)e^{-i{E}_{2}t/\hslash }),\end{array}$$ where we assume the Hamiltonian eigenfunctions ${\varphi }_1(x)$, ${\varphi }_2(x)$ are normalized and real with energy eigenvalues $E_1$, $E_2$. Introducing the frequency $$\begin{array}{}\text{(16.19)}& \omega =\frac{E_2-E_1}{\hslash },\end{array}$$ the probability current is $$\begin{array}{}\text{(16.20)}& \begin{array}{rl}J(x,t)& =\frac{\hslash }{2mi}(\overline{\psi }{\partial }_x\psi -\psi {\partial }_x\overline{\psi })\\ & =\frac{\hslash }{4mi}({\varphi }_1{\partial }_x{\varphi }_1+{\varphi }_2{\partial }_x{\varphi }_2+{\varphi }_1{\partial }_x{\varphi }_2{e}^{-i\omega t}+{\varphi }_2{\partial }_x{\varphi }_1{e}^{i\omega t}-c.c.)\\ & =\frac{\hslash }{2m}({\varphi }_2{\partial }_x{\varphi }_1-{\varphi }_1{\partial }_x{\varphi }_2)\sin (\omega t).\end{array}\end{array}$$ The probability current therefore oscillates in time with frequency $\omega$.

16.5 Example: Plane Waves

Now consider the stationary plane wave solution for a free particle on a line, $$\begin{array}{}\text{(16.21)}& {\psi }_p(x,t)=C{e}^{i(px-E_{p}t)/\hslash },\end{array}$$ where $E_p=p^2/2m$. The probability density and current are constant $$\begin{array}{}\text{(16.22)}& \begin{array}{rl}P(x,t)& =|C{|}^2,\\ J(x,t)& =|C{|}^2\frac{p}{m}.\end{array}\end{array}$$ Note that the probability current is equal to the probability density multiplied by the velocity $p/m$ of the wave. This is consistent with the continuity equation. It evades the statement that $J(x,t)=0$ for a stationary wave function because it is not square-normalisable.

Despite the fact that they are not square-normalizable, plane wave solutions are useful in “scattering problems” in quantum mechanics. This will be the starting point for the next lecture.

16.6 Problems

1. Interpretation of the probability current:

   Let $P_{ab}(t)$ be the probability to find the particle in the interval $a<x<b$.

   1. Write down a definite integral for $P_{ab}(t)$.

   2. Write down the continuity equation and use it to show that $$\begin{array}{}\text{(16.23)}& {\partial }_t{P}_{ab}(t)=J(a,t)-J(b,t).\end{array}$$

   3. Hence discuss the physical interpretation of $J(x,t)$.

   Solution ▶

   1. The probability to find the particle in the interval $a\le x\le b$ is $$\begin{array}{}\text{(16.24)}& P_{ab}(t)={\int }_a^{b}dxP(x,t)={\int }_a^{b}dx|\psi (x,t){|}^2.\end{array}$$

   2. We compute the derivative with respect to time and use the continuity equation $$\begin{array}{}\text{(16.25)}& \begin{array}{rl}{\partial }_t{P}_{ab}(t)& ={\int }_a^{b}dx{\partial }_{t}P(x,t)\\ & =-{\int }_a^{b}dx{\partial }_{x}J(x,t)\\ & =J(a,t)-J(b,t).\end{array}\end{array}$$

   3. This equation represents the conservation of probability: the rate of change of the probability to find the particle in the region $a<x<b$, is equal to the rate the probability is flowing in / out at the boundaries $x=a$ and $x=b$.

2. Stationary probability current:

   Consider a stationary solution of Schrödinger’s equation, $$\begin{array}{}\text{(16.26)}& \psi (x,t)=e^{-iEt/\hslash }\varphi (x).\end{array}$$

   1. Write down an expression for the probability density $P(x,t)$ and show that it is independent of $t$.

   2. Write down an expression for the probability current $J(x,t)$ and show that it is independent of $t$.

   3. Using the continuity equation and part (a), show that $J(x,t)$ is also independent of $x$.

   4. Hence explain why $J(x,t)=0$ if $\varphi (x)$ is square-normalizable.

   Solution ▶

   1. For a stationary wavefunction, the probability density is $$\begin{array}{}\text{(16.27)}& P(x,t)=|\psi (x,t){|}^2=|\varphi (x){|}^2\end{array}$$ which is independent of time.

   2. For a stationary wavefunction, the probability current is $$\begin{array}{}\text{(16.28)}& \begin{array}{rl}J(x,t)& =\frac{\hslash }{2mi}(\overline{\psi }(x,t){\partial }_x\psi (x,t)-\psi (x,t){\partial }_x\overline{\psi }(x,t))\\ & =\frac{\hslash }{2mi}(\overline{\varphi }(x){\partial }_x\varphi (x)-\varphi (x){\partial }_x\overline{\varphi }(x)),\end{array}\end{array}$$ which is independent of time.

   3. The continuity equation is $$\begin{array}{}\text{(16.29)}& {\partial }_{t}P+{\partial }_{x}J=0.\end{array}$$ From part (a), we find ${\partial }_{x}J=0$ and therefore the probability current is also independent of position.

   4. For a square-normalizable wavefunction, $\varphi (x)\to 0$ and therefore $J\to 0$ as $|x|\to \mathrm{\infty }$. Since $J$ is independent of position, $J=0$ everywhere.

3. Probability current in an infinite potential well:

   Consider an infinite potential well $0<x<L$.

   1. By integrating the continuity equation over $0<x<L$, explain why $$\begin{array}{}\text{(16.30)}& J(0,t)-J(L,t)=0.\end{array}$$

   2. Show that the standard boundary conditions on the wavefunction $\psi (x,t)$ at $x=0$ and $x=L$ imply the stronger conditions $$\begin{array}{}\text{(16.31)}& J(0,t)=J(L,t)=0.\end{array}$$

   3. What is the physical interpretation of these results?

   Solution ▶

   1. From the continuity equation ${\partial }_{t}P+{\partial }_{x}J=0$, the time-derivative of the total probability to find the particle anywhere in $0<x<L$ is $$\begin{array}{}\text{(16.32)}& \frac{d}{dt}{\int }_0^{L}Pdx={\int }_0^L{\partial }_{t}Pdx=-{\int }_0^L{\partial }_{x}Jdx=J(0,t)-J(L,t).\end{array}$$ For the conservation of the total probability we therefore require $J(0,t)=J(L,t)$ for all $t$.

   2. The standard boundary condition for the infinite-square well is $$\begin{array}{}\text{(16.33)}& \psi (0,t)=\psi (L,t)=0.\end{array}$$ We do not set the spatial derivative of the wavefunction to vanish at $x=0,L$ because the potential jumps by an infinite amount there. Nevertheless, from the definition of the probability current $$\begin{array}{}\text{(16.34)}& J=\frac{\hslash }{2mi}(\overline{\psi }{\partial }_x\psi -\psi {\partial }_x\overline{\psi })\end{array}$$ we find $$\begin{array}{}\text{(16.35)}& J(0,t)=J(L,t)=0.\end{array}$$ This is stronger than the condition from part (a). Intuitively, the probability to find the particle in the regions $x<0$ and $x>L$ is zero, so there should not be any probability flowing into or our of these regions.

4. TUTORIAL 4
   Time-dependence of the probability current Consider the infinite potential well $0<x<L$ with wavefunction $$\begin{array}{}\text{(16.36)}& \psi (x,t)=\frac{1}{\sqrt{2}}({\varphi }_1(x)e^{-i{E}_{1}t/\hslash }+{\varphi }_2(x)e^{-i{E}_{2}t/\hslash }).\end{array}$$

   1. Show that the probability current has the form $$\begin{array}{}\text{(16.37)}& J(x,t)=C{\sin }^3\left(\frac{\pi x}{L}\right)\sin (\omega t)\end{array}$$ where $\omega =(E_2-E_1)/\hslash$ and $C>0$ is a constant.

   2. Show that $$\begin{array}{}\text{(16.38)}& J(0,t)=J(L,t)=0.\end{array}$$ What is the physical interpretation of this result?

   3. Sketch the probability current $J(x,t)$ at times $$\begin{array}{}\text{(16.39)}& t=0,\frac{\pi }{2\omega },\frac{\pi }{\omega },\frac{3\pi }{2\omega },\frac{2\pi }{\omega }.\end{array}$$

   4. In which direction is the probability “flowing" when $$\begin{array}{}\text{(16.40)}& \text{(i)}0<t<\frac{\pi }{\omega }\text{(ii)}\frac{\pi }{\omega }<t<\frac{2\pi }{\omega }?\end{array}$$

   5. Compare this to a sketch of the expectation value $$\begin{array}{}\text{(16.41)}& ⟨x⟩=\frac{L}{2}-A\cos (\omega t),\end{array}$$ where $0<A<\frac{L}{2}$. Is it consistent?

   Hint 1: For part (a), you may assume the result from section 16.5 of the lecture notes.

   Hint 2: For part (a), you may use the trigonometric identity $$\begin{array}{}\text{(16.42)}& 2{\sin }^{3}y=\sin (2y)\cos (y)-2\cos (2y)\sin (y).\end{array}$$

   Solution ▶

   1. In the section “Example: Sum of Two Stationary Wave functions”, we derived the following formula for the probability current of a sum of Hamiltonian eigenfunctions of energies $E_1$ and $E_2$, $$\begin{array}{}\text{(16.43)}& J(x,t)=\frac{\hslash }{2m}({\varphi }_2{\partial }_x{\varphi }_1-{\varphi }_1{\partial }_x{\varphi }_2)\sin (\omega t)\end{array}$$ where $\omega =(E_2-E_1)/\hslash$. In the present case, $$\begin{array}{}\text{(16.44)}& {\varphi }_1=\sqrt{\frac{2}{L}}\sin \left(\frac{\pi x}{L}\right){\varphi }_2=\sqrt{\frac{2}{L}}\sin \left(\frac{2\pi x}{L}\right)\end{array}$$ and hence $$\begin{array}{}\text{(16.45)}& \begin{array}{rl}J(x,t)& =\frac{\hslash }{2m}({\varphi }_2{\partial }_x{\varphi }_1-{\varphi }_1{\partial }_x{\varphi }_2)\sin (\omega t)\\ & =\frac{\hslash \pi }{m{L}^2}(\sin \left(\frac{2\pi x}{L}\right)\cos \left(\frac{\pi x}{L}\right)-2\cos \left(\frac{2\pi x}{L}\right)\sin \left(\frac{\pi x}{L}\right))\sin (\omega t)\\ & =\frac{2\hslash \pi }{m{L}^2}{\sin }^3\left(\frac{\pi x}{L}\right)\sin (\omega t)\end{array}\end{array}$$ using the hint.

   2. It is clear that $J(0,t)=J(L,t)=0$ since $\sin (0)=\sin (\pi )=0$. The physical explanation is that the wavefunction vanishes for $x<0$ and $x>L$ so there cannot be any probability flowing into / out of these regions.

   3. The sequence of sketches is

      

   4. 1. For $0<t<\frac{\pi }{\omega }$, $J>0$ so the probability is flowing to the right.

      2. For $\frac{\pi }{\omega }<t<\frac{2\pi }{\omega }$, $J<0$ so the probability is flowing to the left.

   5. Let us compare this to the sketch below of the expectation value $$\begin{array}{}\text{(16.46)}& ⟨x⟩=\frac{L}{2}-A\cos (\omega t)\end{array}$$ where $A=16L/9{\pi }^2$.

      

      We see that whenever $J>0$ the position expectation value is moving to the right and vice verse, as expected.