# Question 70111

It follows from integrating the Schrödinger equation for $| \setminus \varphi \left(x , t\right) {|}^{2} = \setminus \varphi \left(x , t\right) \setminus \overline{\setminus \varphi} \left(x , t\right)$.

#### Explanation:

Let $\setminus \varphi \left(x , t\right)$ be a (normalizable) solution of the Schrödinger equation:

iħ\frac{\partial}{\partial t}\varphi(x,t) = -\frac{ħ^{2}}{2m}\frac{\partial^{2}}{\partial x^{2}}\varphi(x,t) + V(x)\varphi(x,t),

and let $\setminus \overline{\setminus \varphi} \left(x , t\right)$ denote it's complex conjugate. Then, taking the complex conjugate of the Schrödinger equation (check all the calculations!), the following equation is satisfied for $\setminus \overline{\setminus \varphi} \left(x , t\right)$:

-iħ\frac{\partial}{\partial t}\bar{\varphi}(x,t) = -\frac{ħ^{2}}{2m}\frac{\partial^{2}}{\partial x^{2}}\bar{\varphi}(x,t) + V(x)\bar{\varphi}(x,t).

We also have the standard result

$\setminus \varphi \left(x , t\right) \setminus \overline{\setminus \varphi} \left(x , t\right) = | \setminus \varphi \left(x , t\right) {|}^{2} .$

Taking the time derivative and multiplying by iħ, we get:

iħ\frac{\partial}{\partial t}|\varphi(x,t)|^{2} = iħ\frac{\partial}{\partial t}[\varphi(x,t)\bar{\varphi}(x,t)]
=\bar{\varphi}(x,t)iħ\frac{\partial}{\partial t}\varphi(x,t) + \varphi(x,t)iħ\frac{\partial}{\partial t}\bar{\varphi}(x,t).

Plugging in the two versions of the Schrödinger equation presented above:

iħ\frac{\partial}{\partial t}|\varphi(x,t)|^{2} = -\frac{ħ^{2}}{2m}[\bar{\varphi}(x,t)(\frac{\partial^{2}}{\partial x^{2}}+V(x))\varphi(x,t)
- \varphi(x,t)(\frac{\partial^{2}}{\partial x^{2}}+V(x))\bar{\varphi}(x,t)].

The terms with the potential $V \left(x\right)$ clearly cancel. Therefore,

iħ\frac{\partial}{\partial t}|\varphi(x,t)|^{2} = -\bar{\varphi}(x,t)\frac{ħ^{2}}{2m}\frac{\partial^{2}}{\partial x^{2}}\varphi(x,t)
+ \varphi(x,t)\frac{ħ^{2}}{2m}\frac{\partial^{2}}{\partial x^{2}}\bar{\varphi}(x,t).

Let's define a convenient quantity:

j(x,t) = \frac{iħ}{2m}[\varphi(x,t)\frac{\partial}{\partial x}\bar{\varphi}(x,t) - \bar{\varphi}(x,t)\frac{\partial}{\partial x}\varphi(x,t)].

We'll call $j$ the probability current, in analogy with the probability density $\setminus \rho \left(x , t\right) = | \setminus \varphi \left(x , t\right) {|}^{2}$. The motivation behind this definition will be clear in what follows.

Using the chain rule once again,

\frac{\partial}{\partial x}j(x,t) = \frac{iħ}{2m}[-\bar{\varphi}(x,t)\frac{\partial^{2}}{\partial x^{2}}\varphi(x,t) + \varphi(x,t)\frac{\partial^{2}}{\partial x^{2}}\bar{\varphi}(x,t)
- \frac{\partial}{\partial x}\bar{\varphi}(x,t)\frac{\partial}{\partial x}\varphi(x,t) + \frac{\partial}{\partial x}\varphi(x,t)\frac{\partial}{\partial x}\bar{\varphi}(x,t)]
=\frac{iħ}{2m}[-\bar{\varphi}(x,t)\frac{\partial^{2}}{\partial x^{2}}\varphi(x,t) + \varphi(x,t)\frac{\partial^{2}}{\partial x^{2}}\bar{\varphi}(x,t)].

Now we can see, from our previous result, that

iħ\frac{\partial}{\partial t}|\varphi(x,t)|^{2} = -iħ\frac{\partial}{\partial x}j(x,t).

Dividing this equation by iħ and writing it in terms of $\setminus \rho$ and $j$, we get

$\setminus \frac{\setminus \partial}{\setminus \partial t} \setminus \rho \left(x , t\right) + \setminus \frac{\setminus \partial}{\setminus \partial x} j \left(x , t\right) = 0 ,$

wich is just the continuity equation. This justifies our definition of probability current.

This allows us to write:

$\setminus \frac{d}{\mathrm{dt}} \setminus {\int}_{- \infty}^{+ \infty} | \setminus \varphi \left(x , t\right) {|}^{2} \mathrm{dx} = \setminus {\int}_{- \infty}^{+ \infty} \setminus \frac{\setminus \partial}{\setminus \partial t} | \setminus \varphi \left(x , t\right) {|}^{2} \mathrm{dx}$
$= \setminus {\int}_{- \infty}^{+ \infty} \setminus \frac{\setminus \partial}{\setminus \partial t} \setminus \rho \left(x , t\right) \mathrm{dx} = - \setminus {\int}_{- \infty}^{+ \infty} \setminus \frac{\setminus \partial}{\setminus \partial x} j \left(x , t\right) \mathrm{dx}$
$= \setminus {\lim}_{x \setminus \to + \setminus \infty} j \left(x , t\right) - \setminus {\lim}_{x \setminus \to - \setminus \infty} j \left(x , t\right) .$

A normalizable wavefunction must clearly vanish at infinity. Therefore,

$\setminus {\lim}_{x \setminus \to + \setminus \infty} j \left(x , t\right) = \setminus {\lim}_{x \setminus \to - \setminus \infty} j \left(x , t\right) = 0 ,$

and we have proved that

$\setminus \frac{d}{\mathrm{dt}} \setminus {\int}_{- \infty}^{+ \infty} | \setminus \varphi \left(x , t\right) {|}^{2} \mathrm{dx} = 0 ,$

that is, the norm of the wavefunction does not vary with time. For this reason, we say that, in Quantum Mechanics, time evolution is unitary.

This result can be generalized to three (in fact, any number of) dimensions:

$\setminus \frac{\setminus \partial}{\setminus \partial t} \setminus \rho \left(\setminus \vec{r} , t\right) + \setminus \nabla \cdot \setminus \vec{j} \left(\setminus \vec{r} , t\right) = 0 ,$

where

$\setminus \rho \left(\setminus \vec{r} , t\right) = | \setminus \varphi \left(\setminus \vec{r} , t\right) {|}^{2} ,$
j(\vec{r},t) = \frac{iħ}{2m}[\varphi(\vec{r},t)\grad\bar{\varphi}(\vec{r},t) - \bar{\varphi}(\vec{r},t)\grad\varphi(\vec{r},t)].

The probability current can also be writen in terms of the momentum operator \hat{P}=-iħ\grad# (in the position representation):

$j \left(\setminus \vec{r} , t\right) = \setminus \frac{1}{2 m} \left[\setminus \varphi \left(\setminus \vec{r} , t\right) \setminus \hat{P} \setminus \overline{\setminus \varphi} \left(\setminus \vec{r} , t\right) - \setminus \overline{\setminus \varphi} \left(\setminus \vec{r} , t\right) \setminus \hat{P} \setminus \varphi \left(\setminus \vec{r} , t\right)\right]$
$= \setminus \frac{1}{m} \text{Re} \left[\setminus \varphi \left(\setminus \vec{r} , t\right) \setminus \hat{P} \setminus \overline{\setminus \varphi} \left(\setminus \vec{r} , t\right)\right]$.

Note also that we have assumed that the potential $V$ is a real function, so that $\setminus \overline{V} = V$. If we remove this hypothesis (that is, we let $V$ be a complex function), the result does not hold. This is useful when considering particle decay, a situation where we expect the probability density to decrease exponentially with time (this decrease depends on the imaginary part of $V$).

A further generalization can be obtained with the Dirac (or bra-ket) notation:

$\setminus \frac{d}{\mathrm{dt}} | | \setminus \varphi | {|}^{2} = \setminus \frac{d}{\mathrm{dt}} \setminus \left\langle \setminus \varphi | \setminus \varphi \setminus\right\rangle = 0 ,$

wich holds for very general quantum mechanical systems.