16 The Continuity Equation

16.1 The Continuity Equation

Our starting point is the probability density \[P(x,t) := \big|\psi(x,t)\big|^2 \,.\] 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, \[\begin{aligned} \partial_t \psi(x,t) & = -\frac{i}{\hbar}\left( - \frac{\hbar^2}{2m} \partial_x^2 + V(x) \right) \psi(x,t) \, , \\[1ex] \partial_t \overline{\psi}(x,t) & = \frac{i}{\hbar}\left( - \frac{\hbar^2}{2m} \partial_x^2 + V(x) \right) \overline{\psi}(x,t) \, , \end{aligned}\] where we recall that the potential \(V(x)\) is a real analytic function. We can now compute the time derivative of the probability density, \[\begin{aligned} \partial_t P & = \partial_t |\psi|^2 \\[1ex] & = \psi \, ( \partial_t \overline{\psi} ) + \overline{\psi} \, ( \partial_t \psi ) \\[1ex] & =\frac{i}{\hbar} \psi \left(- \frac{\hbar^2}{2m} \partial_x^2+V \right)\overline\psi - \frac{i}{\hbar} \overline\psi \left(- \frac{\hbar^2}{2m} \partial_x^2 +V \right)\psi \, . \end{aligned}\] Note that the dependence on the potential \(V(x)\) cancels out between the two terms and the remainder becomes \[\begin{aligned} \partial_t P & = \frac{\hbar}{2mi}\left( \psi \, \partial_x^2\overline\psi - \bar\psi \, \partial^2_x\psi \right) \\[1ex] & = \frac{\hbar}{2mi} \partial_x \left( \psi \partial_x \overline\psi - \overline\psi \partial_x \psi \right) \\[1ex] & = - \partial_x J \, , \end{aligned}\] where the symbol \(J\), \[J := \frac{\hbar}{2mi}( \bar\psi \partial_x \psi - \psi \partial_x \bar\psi )\] is known as the “probability current density”. The result \[\partial_t P + \partial_x J = 0 \,,\] 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 \[P_{ab}(t) = \int^b_aP(x,t) \,{\rm d}x\] 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, \[\begin{aligned} \frac{d}{dt} P_{ab}(t) & = \int^b_a \partial_t P(t,x) \,{\rm d}x \\[1ex] & = - \int^b_a \partial_x J(t,x) \,{\rm d}x \\[1ex] & = J(a,t) - J(b,t) \, . \end{aligned}\]

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 \(a \to -\infty\) and \(b\to \infty\), the equation becomes \[\frac{d}{dt} \int^\infty_{-\infty} P(x,t) \,{\rm d}x = 0\, ,\] since if the wave function is normalizable then \(\psi(x,t) \to 0\) and hence \(J(x,t) \to 0\) as \(x \to \pm \infty\). We therefore recover the conservation of the total probability to find the probability. In particular, if the wave function is normalised at \(t=t_0\), \[\int^\infty_{-\infty} P(x, t=t_0) \,{\rm d}x = 1\,,\] 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 \(b = x+ \,{\rm d}x\) we find \[\partial_t P(x,t) \,{\rm d}x = J(x,t) - J(x+{\rm d}x,t) \, .\] In the limit \({\rm d}x \to 0\), we recover the continuity equation \(\partial_tP(x,t) + \partial_xJ(x,t)\), 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, \[\psi(x,t) = \phi(x) e^{-iEt/\hbar}\,, \qquad \hat{H}\,\phi(x) = E \phi(x) \, .\] The probability density and current are independent of time, \[\begin{aligned} P(x,t) & = |\phi(x)|^2\,, \\[1ex] J (x,t) & = \frac{\hbar}{2mi}\big( \bar\phi(x) \partial_x \phi(x) - \phi(x)\partial_x \bar\phi(x) \big) \,. \end{aligned}\] The continuity equation tells us that \(\partial_xJ(x,t) = 0\) and therefore \(J(x,t) = J_0\) is constant. Furthermore, if \(\phi(x)\) is square-normalizable then \(J(x,t) \to 0\) as \(x \to \pm \infty\). 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 \(0<x<L\), \[\phi(x) = \sqrt{\frac{2}{L}}\sin\left(\frac{n\pi x}{L}\right) \,,\] for some \(n \in \mathbb{Z}_{>0}\). Since \(\phi(x)\) is a real function, \[\begin{aligned} J & = \frac{\hbar}{2mi}( \bar\psi \partial_x \psi - \psi \partial_x \bar\psi ) = \frac{\hbar}{2mi}( \phi \partial_x \phi - \phi \partial_x \phi ) = 0 \, , \end{aligned}\] 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, \[\psi(x,t) = \frac{1}{\sqrt{2}}\left( \phi_1(x) e^{-iE_1t/\hbar} + \phi_2(x) e^{-iE_2t/\hbar} \right) \, ,\] where we assume the Hamiltonian eigenfunctions \(\phi_1(x)\), \(\phi_2(x)\) are normalized and real with energy eigenvalues \(E_1\), \(E_2\). Introducing the frequency \[\omega = \frac{E_2-E_1}{\hbar} \, ,\] the probability current is \[\begin{aligned} J(x,t) & = \frac{\hbar}{2mi} \left( \bar\psi \, \partial_x \psi - \psi \, \partial_x \bar\psi \right) \\[1ex] & = \frac{\hbar}{4mi} \left( \phi_1\partial_x\phi_1 + \phi_2\partial_x\phi_2 + \phi_1\partial_x\phi_2 \, e^{-i\omega t} + \phi_2\partial_x\phi_1 \, e^{i\omega t} - c.c. \right) \\[1ex] & = \frac{\hbar}{2m}\left(\phi_2\partial_x\phi_1 - \phi_1\partial_x\phi_2 \right) \sin(\omega t) \, . \end{aligned}\] 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, \[\psi_p(x,t) = C \, e^{i (p x- E_pt )/ \hbar} \, ,\] where \(E_p = p^2/2m\). The probability density and current are constant \[\begin{aligned} P(x,t) & = |C|^2\,, \\[1ex] J(x,t) & = |C|^2 \frac{p}{m}\,. \end{aligned}\] 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 \[\partial_t P_{ab}(t) = J(a,t) - J(b,t) \, .\]

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

   Solution ▶

   1. The probability to find the particle in the interval \(a\leq x\leq b\) is \[P_{ab}(t) = \int^b_a dx \, P(x,t) = \int^b_a dx \, |\psi(x,t)|^2 \, .\]

   2. We compute the derivative with respect to time and use the continuity equation \[\begin{aligned} \partial_t P_{ab}(t) & = \int^b_a dx \, \partial_t P(x,t) \\ \nonumber & = - \int^b_a dx \, \partial_xJ(x,t) \\ & = J(a,t) - J(b,t) \, . \end{aligned}\]

   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, \[\psi(x,t) = e^{- i E t / \hbar} \phi(x) \, .\]

   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 \(\phi(x)\) is square-normalizable.

   Solution ▶

   1. For a stationary wavefunction, the probability density is \[P(x,t) =|\psi(x,t)|^2 = |\phi(x)|^2\] which is independent of time.

   2. For a stationary wavefunction, the probability current is \[\begin{aligned} \nonumber J(x,t) & = \frac{\hbar}{2mi}(\bar\psi(x,t) \partial_x \psi(x,t) - \psi(x,t) \partial_x \bar\psi(x,t)) \\ & = \frac{\hbar}{2mi}(\bar\phi(x) \partial_x \phi(x) - \phi(x) \partial_x \bar\phi(x)) \, , \end{aligned}\] which is independent of time.

   3. The continuity equation is \[\partial_tP +\partial_x J = 0\, .\] From part (a), we find \(\partial_xJ = 0\) and therefore the probability current is also independent of position.

   4. For a square-normalizable wavefunction, \(\phi(x) \to 0\) and therefore \(J \to 0\) as \(|x| \to \infty\). 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 \[J(0,t) - J(L,t) = 0 \, .\]

   2. Show that the standard boundary conditions on the wavefunction \(\psi(x,t)\) at \(x = 0\) and \(x=L\) imply the stronger conditions \[J(0,t) = J(L,t) = 0 \, .\]

   3. What is the physical interpretation of these results?

   Solution ▶

   1. From the continuity equation \(\partial_t P + \partial_xJ = 0\), the time-derivative of the total probability to find the particle anywhere in \(0<x<L\) is \[\frac{d}{dt} \int^L_0P dx = \int^L_0 \partial_t P dx = - \int^L_0 \partial_xJ dx = J(0,t) - J(L,t) \, .\] 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 \[\psi(0,t) = \psi(L,t) = 0 \, .\] 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 \[J = \frac{\hbar}{2mi}(\bar\psi \partial_x \psi - \psi \partial_x \bar\psi)\] we find \[J(0,t) = J(L,t) = 0\, .\] 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 \[\psi(x,t) = \frac{1}{\sqrt{2}}(\phi_1(x)e^{-iE_1t/\hbar}+\phi_2(x)e^{-iE_2t/\hbar}) \, .\]

   1. Show that the probability current has the form \[J(x,t) = C \sin^3\left(\frac{\pi x}{L} \right) \sin(\omega t)\] where \(\omega = (E_2-E_1) / \hbar\) and \(C>0\) is a constant.

   2. Show that \[J(0,t) = J(L,t) = 0 \, .\] What is the physical interpretation of this result?

   3. Sketch the probability current \(J(x,t)\) at times \[t = 0, \, \frac{\pi}{2\omega}, \, \frac{\pi}{\omega} , \, \frac{3\pi}{2\omega}, \, \frac{2\pi}{\omega} \, .\]

   4. In which direction is the probability “flowing" when \[\text{(i)} \quad 0<t<\frac{\pi}{\omega} \quad\qquad \text{(ii)} \quad \frac{\pi}{\omega} < t < \frac{2\pi}{\omega} \, ?\]

   5. Compare this to a sketch of the expectation value \[\langle x\rangle = \frac{L}{2} - A \cos(\omega t) \, ,\] 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 \[2\sin^3y = \sin(2y)\cos(y) - 2 \cos(2y)\sin(y) \, .\]

   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 \(E_2\), \[J(x,t) = \frac{\hbar}{2m}\left( \phi_2\partial_x\phi_1 - \phi_1\partial_x \phi_2\right)\sin(\omega t)\] where \(\omega = (E_2-E_1)/\hbar\). In the present case, \[\phi_1 = \sqrt{\frac{2}{L}}\sin\left(\frac{\pi x}{L}\right) \qquad \phi_2 = \sqrt{\frac{2}{L}}\sin\left(\frac{2\pi x}{L}\right)\] and hence \[\begin{aligned} J(x,t) & = \frac{\hbar}{2m}\left( \phi_2\partial_x\phi_1 - \phi_1\partial_x \phi_2\right)\sin(\omega t) \\ \nonumber & = \frac{\hbar\pi}{mL^2} \left( \sin\left(\frac{2\pi x}{L}\right) \cos\left(\frac{\pi x}{L}\right) - 2\cos\left(\frac{2\pi x}{L}\right)\sin\left(\frac{\pi x}{L}\right)\right)\sin(\omega t) \\ & = \frac{2\hbar\pi}{mL^2} \sin^3\left(\frac{\pi x}{L}\right)\sin(\omega t) \end{aligned}\] 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 \[\langle x \rangle = \frac{L}{2} - A \cos(\omega t)\] 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.