Probability current

Required math: calculus

Required physics: Schrödinger equation

Reference: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 1.14.

Using the Schrödinger equation we can derive an interesting quantity called the probability current. Using the probabilistic interpretation of the wave function, the probability of a particle being between {x=a} and {x=b} is

\displaystyle  P_{ab}=\int_{a}^{b}\left|\Psi^*\Psi\right|dx \ \ \ \ \ (1)

The rate of change of this probability can then be expressed in terms of spatial derivatives using the Schrödinger equation:

\displaystyle   \frac{dP_{ab}}{dt} \displaystyle  = \displaystyle  \int_{a}^{b}\left[\frac{\partial\Psi^*}{\partial t}\Psi+\frac{\partial\Psi}{\partial t}\Psi^*\right]dx\ \ \ \ \ (2)
\displaystyle  \displaystyle  = \displaystyle  \int_{a}^{b}\left\{ -\frac{i\hbar}{2m}\frac{\partial^{2}\Psi^*}{\partial x^{2}}-\frac{1}{i\hbar}V\Psi^*\right\} \Psi dx\ \ \ \ \ (3)
\displaystyle  \displaystyle  \displaystyle  +\int_{a}^{b}\left\{ \frac{i\hbar}{2m}\frac{\partial^{2}\Psi}{\partial x^{2}}+\frac{1}{i\hbar}V\Psi\right\} \Psi^*dx\ \ \ \ \ (4)
\displaystyle  \displaystyle  = \displaystyle  \frac{i\hbar}{2m}\int_{a}^{b}\left[\frac{\partial^{2}\Psi}{\partial x^{2}}\Psi^*-\frac{\partial^{2}\Psi^*}{\partial x^{2}}\Psi\right]dx \ \ \ \ \ (5)

We can now apply integration by parts to each term.

\displaystyle   \int_{a}^{b}\frac{\partial^{2}\Psi}{\partial x^{2}}\Psi^*dx \displaystyle  = \displaystyle  \left.\frac{\partial\Psi}{\partial x}\Psi^*\right|_{a}^{b}-\int_{a}^{b}\frac{\partial\Psi}{\partial x}\frac{\partial\Psi^*}{\partial x}dx\ \ \ \ \ (6)
\displaystyle  -\int_{a}^{b}\frac{\partial^{2}\Psi^*}{\partial x^{2}}\Psi dx \displaystyle  = \displaystyle  \left.-\frac{\partial\Psi^*}{\partial x}\Psi\right|_{a}^{b}+\int_{a}^{b}\frac{\partial\Psi}{\partial x}\frac{\partial\Psi^*}{\partial x}dx \ \ \ \ \ (7)

Adding these terms together, we get

\displaystyle  \frac{dP_{ab}}{dt}=\frac{i\hbar}{2m}\left[\left.\frac{\partial\Psi}{\partial x}\Psi^*\right|_{a}^{b}\left.-\frac{\partial\Psi^*}{\partial x}\Psi\right|_{a}^{b}\right] \ \ \ \ \ (8)

If we define the probability current as

\displaystyle  J(x,t)\equiv\frac{i\hbar}{2m}\left(\frac{\partial\Psi^*}{\partial x}\Psi-\frac{\partial\Psi}{\partial x}\Psi^*\right) \ \ \ \ \ (9)



we can write the rate of change of probability as

\displaystyle  \frac{dP_{ab}}{dt}=J(a,t)-J(b,t) \ \ \ \ \ (10)

As an example, if the wave function is given by

\displaystyle  \Psi(x,t)=\left(\frac{2am}{\pi\hbar}\right)^{1/4}e^{-a\left[\left(mx^{2}/\hbar\right)+it\right]} \ \ \ \ \ (11)



(we’ve taken the constant in front so that it normalizes the wave function), then

\displaystyle  \frac{\partial\Psi}{\partial x}=-\frac{1}{\pi^{1/4}}\left(\frac{2am}{\hbar}\right)^{5/4}xe^{-a\left[\left(mx^{2}/\hbar\right)+it\right]} \ \ \ \ \ (12)

So for the probability current we get

\displaystyle   J(x,t) \displaystyle  = \displaystyle  -\frac{i\hbar}{2m}\frac{1}{\pi^{1/4}}\left(\frac{2am}{\hbar}\right)^{5/4}xe^{-a\left[\left(mx^{2}/\hbar\right)-it\right]}\left(\frac{2am}{\pi\hbar}\right)^{1/4}e^{-a\left[\left(mx^{2}/\hbar\right)+it\right]}\ \ \ \ \ (13)
\displaystyle  \displaystyle  \displaystyle  +\frac{i\hbar}{2m}\frac{1}{\pi^{1/4}}\left(\frac{2am}{\hbar}\right)^{5/4}xe^{-a\left[\left(mx^{2}/\hbar\right)+it\right]}\left(\frac{2am}{\pi\hbar}\right)^{1/4}e^{-a\left[\left(mx^{2}/\hbar\right)-it\right]}\ \ \ \ \ (14)
\displaystyle  \displaystyle  = \displaystyle  0 \ \ \ \ \ (15)

A bit of an anti-climax after all that mathematics.

2 thoughts on “Probability current

  1. Pingback: The free particle: probability current « Physics tutorials

  2. Pingback: Probability current in 3-d « Physics tutorials

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