Physics Friday 57

Consider a single particle wavefunction (in the position basis) $\text{\Psi}(\mathbf{\vec{r}},t)$ . Then for a volume V, the probability P that the particle will be measured to be in V is
$P(t)=\int_{V}\left|\text{\Psi}\right|^2\,dV$ .
The time derivative of this is:
$\begin{eqnarray}\frac{dP}{dt}&=&\frac{d}{dt}\int_{V}\left|\text{\Psi}\right|^2\,dV\\&=&\int_{V}\frac{d}{dt}\left|\text{\Psi}\right|^2\,dV\end{eqnarray}$ .

Now, we recall that $\left|\text{\Psi}\right|^2=\text{\Psi}^*\text{\Psi}$ , where ^* indicates the complex conjugate. Thus, via the product rule,
$\begin{eqnarray}\frac{d}{dt}\left|\text{\Psi}\right|^2&=&\frac{d}{dt}\left(\text{\Psi}^*\text{\Psi}\right)\\&=&\frac{\partial\text{\Psi}^*}{\partial{t}}\text{\Psi}+\text{\Psi}^*\frac{\partial\text{\Psi}}{\partial{t}}\end{eqnarray}$
So
$\frac{dP}{dt}=\int_{V}\frac{\partial\text{\Psi}^*}{\partial{t}}\text{\Psi}+\text{\Psi}^*\frac{\partial\text{\Psi}}{\partial{t}}\,dV$

Now, consider the time dependent Schrödinger equation:
$\imath\hbar\frac{\partial\text{\Psi}}{\partial{t}}=\hat{H}\text{\Psi}=\frac{-\hbar^2}{2m}\nabla^2\text{\Psi}+U\text{\Psi}$
Solving for the time derivative, we get
$\frac{\partial\text{\Psi}}{\partial{t}}=\frac{\imath\hbar}{2m}\nabla^2\text{\Psi}-\frac{\imath}{\hbar}U\text{\Psi}$
And taking the complex conjugate of that:
$\frac{\partial\text{\Psi}^*}{\partial{t}}=-\frac{\imath\hbar}{2m}\nabla^2\text{\Psi}^*+\frac{\imath}{\hbar}U\text{\Psi}^*$ .
(Note that the potential $U(\mathbf{\vec{r}})$ is real).
Thus
$\frac{\partial\text{\Psi}^*}{\partial{t}}\text{\Psi}=-\frac{\imath\hbar}{2m}\text{\Psi}\nabla^2\text{\Psi}^*+\frac{\imath}{\hbar}U\text{\Psi}^*\text{\Psi}$ ,
and
$\text{\Psi}^*\frac{\partial\text{\Psi}}{\partial{t}}=\frac{\imath\hbar}{2m}\text{\Psi}^*\nabla^2\text{\Psi}-\frac{\imath}{\hbar}U\text{\Psi}^*\text{\Psi}$
Adding these, we get
$\frac{\partial\text{\Psi}^*}{\partial{t}}\text{\Psi}+\text{\Psi}^*\frac{\partial\text{\Psi}}{\partial{t}}=-\frac{\hbar}{2m\imath}\left[\text{\Psi}^*\nabla^2\text{\Psi}-\text{\Psi}\nabla^2\text{\Psi}^*\right]$
(the potential energy terms cancel).
So
$\frac{dP}{dt}=-\int_{V}\frac{\hbar}{2m\imath}\left[\text{\Psi}^*\nabla^2\text{\Psi}-\text{\Psi}\nabla^2\text{\Psi}^*\right]\,dV$ .

Here, we need to use some vector calculus; namely, the product rule for the divergence operator $\vec{\nabla}\cdot$ : for scalar valued function φ and vector field $\mathbf{\vec{F}}$ , the divergence of their product is given by
$\vec{\nabla}\cdot(\phi\mathbf{\vec{F}})=\vec{\nabla}\phi\cdot\mathbf{\vec{F}}+\phi\vec{\nabla}\cdot\mathbf{\vec{F}}$
Now, if our vector field is itself the gradient of a scalar function ψ, then
$\vec{\nabla}\cdot(\phi\vec{\nabla}\psi)=\vec{\nabla}\phi\cdot\vec{\nabla}\psi+\phi\nabla^2\psi$
Swapping φ and ψ,
$\vec{\nabla}\cdot(\psi\vec{\nabla}\phi)=\vec{\nabla}\psi\cdot\vec{\nabla}\phi+\psi\nabla^2\phi$
And taking the difference, we find
$\vec{\nabla}\cdot(\psi\vec{\nabla}\phi-\phi\vec{\nabla}\psi)=\psi\nabla^2\phi-\phi\nabla^2\psi$ .
Putting in our wavefunction and it’s conjugate for φ and ψ,
$\text{\Psi}^*\nabla^2\text{\Psi}-\text{\Psi}\nabla^2\text{\Psi}^*=\vec{\nabla}\cdot\left(\text{\Psi}^*\vec{\nabla}\text{\Psi}-\text{\Psi}\vec{\nabla}\text{\Psi}^*\right)$
So
$\frac{dP}{dt}=-\int_{V}\vec{\nabla}\cdot\left[\frac{\hbar}{2m\imath}\left(\text{\Psi}^*\vec{\nabla}\text{\Psi}-\text{\Psi}\vec{\nabla}\text{\Psi}^*\right)\right]\,dV$ .
Let us call the vector-valued function that is the argument of the divergence in the integrand $\mathbf{\vec{J}}(\mathbf{\vec{r}},t)$ :
$\begin{eqnarray}\mathbf{\vec{J}}&\equiv&\frac{\hbar}{2m\imath}\left(\text{\Psi}^*\vec{\nabla}\text{\Psi}-\text{\Psi}\vec{\nabla}\text{\Psi}^*\right)\\&=&\frac{\hbar}{2m\imath}\left(\text{\Psi}^*\vec{\nabla}\text{\Psi}-\left(\text{\Psi}^*\vec{\nabla}\text{\Psi}\right)^*\right)\\&=&\frac{\hbar}{m}\Im\left(\text{\Psi}^*\vec{\nabla}\text{\Psi}\right)\end{eqnarray}$
Then
$\frac{dP}{dt}=-\int_{V}\vec{\nabla}\cdot\mathbf{\vec{J}}\,dV$
Now, recalling that the probability P is the integral over V of the probability density $\left|\text{\Psi}\right|^2$
So in terms of the probability density $p(\mathbf{\vec{r}},t)\equiv\left|\text{\Psi}\right|^2$
$\begin{eqnarray}\int_{V}\frac{\partial{p}}{\partial{t}}\,dV&=&-\int_{V}\vec{\nabla}\cdot\mathbf{\vec{J}}\,dV\\\int_{V}\frac{\partial{p}}{\partial{t}}\,dV+\int_{V}\vec{\nabla}\cdot\mathbf{\vec{J}}\,dV&=&{0}\\\int_{V}\frac{\partial{p}}{\partial{t}}+\vec{\nabla}\cdot\mathbf{\vec{J}}\,dV&=&{0}\end{eqnarray}$
and as this holds for all V, the integrand must vanish:
$\frac{\partial{p}}{\partial{t}}+\vec{\nabla}\cdot\mathbf{\vec{J}}=0$

Now, this should look familiar to some of you. For any conserved quantity ρ with a flux given by the function $\vec{f}$ , and no sources or sinks, the quantity and flux obey the continuity equation
$\frac{\partial\rho}{\partial{t}}+\vec{\nabla}\cdot\vec{f}=0$ .
[For example, if ρ is electric charge density, then the conservation of electric charge gives
$\frac{\partial\rho}{\partial{t}}+\vec{\nabla}\cdot\mathbf{\vec{J}}=0$ ,
where $\mathbf{\vec{J}}$ is the electric current density.]
Now, as the probability density for the particle must always integrate over all space to unity, we similarly expect it to be a conserved quantity. Thus the above result is our continuity equation for quantum probability, and so $\mathbf{\vec{J}}(\mathbf{\vec{r}},t)$ as defined above is our probability current (or probability flux). It has units of probability/(area × time), or probability density times velocity.

Note that the continuity equation tells us that for a stationary state, the divergence of the probability current must be zero. However, this does not mean that the current itself must be zero. Consider the three-dimensional plane wave $\text{\psi}(\mathbf{\vec{r}},t)=Ae^{\imath(\mathbf{\vec{k}}\cdot\mathbf{\vec{r}}-\omega{t})}$ .
Then
$\vec{\nabla}\text{\Psi}=\imath\mathbf{\vec{k}}Ae^{\imath(\mathbf{\vec{k}}\cdot\mathbf{\vec{r}}-\omega{t})}=\imath\mathbf{\vec{k}}\text{\Psi}$
$\vec{\nabla}\text{\Psi}^*=-\imath\mathbf{\vec{k}}\text{\Psi}^*$
So
$\begin{eqnarray}\text{\Psi}^*\vec{\nabla}\text{\Psi}-\text{\Psi}\vec{\nabla}\text{\Psi}^*&=&\text{\Psi}^*(\imath\mathbf{\vec{k}}\text{\Psi})-\text{\Psi}(-\imath\mathbf{\vec{k}}\text{\Psi}^*)\\&=&2\imath\mathbf{\vec{k}}\text{\Psi}^*\text{\Psi}\\&=&2\imath|A|^2\mathbf{\vec{k}}\end{eqnarray}$
So
$\begin{eqnarray}\mathbf{\vec{J}}&=&\frac{\hbar}{2m\imath}\left(2\imath|A|^2\mathbf{\vec{k}}\right)\\&=&|A|^2\frac{\hbar\mathbf{\vec{k}}}{m}\end{eqnarray}$ .
Note that as $\frac{\hbar\mathbf{\vec{k}}}{m}=\frac{\mathbf{\vec{p}}}{m}=\mathbf{\vec{v}}$ is the particle’s velocity, the probability current of the plane wave, a stationary state, is the amplitude squared times the particle velocity.

Lastly, consider a wavefunction which has the same complex phase for all locations at any given time; that is $\text{\Psi}(\mathbf{\vec{r}},t)=c(t)\psi(\mathbf{\vec{r}},t)$ , where $\psi(\mathbf{\vec{r}},t)$ is a real-valued function. Then $\text{\Psi}^*=c^*\psi$ , and $\text{\Psi}^*\vec{\nabla}\text{\Psi}=|c|^2\psi\vec{\nabla}\psi=\text{\Psi}\vec{\nabla}\text{\Psi}^*$ , and so we see the probability current is zero for such wavefunctions (one example of which are solutions to the particle in a one-dimensional box).

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Tags: Continuity Equation, Friday Physics, physics, Plane Wave, Probability, Probability Current, Probability Density, Probability Flux, Quantum Mechanics, Schrödinger Equation

This entry was posted on January 30, 2009 at 12:01 am and is filed under Math/Science. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

One Response to “Physics Friday 57”

Physics Friday 121 « Twisted One 151's Weblog Says:
May 31, 2010 at 3:24 pm | Reply
[...] with the conservation of charge in relation to current. The continuity equation is (as mentioned here). For a steady-state magnetic system, we need no change in the net charge density anywhere, and [...]

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