Relation Between the Wave Function and Probability Density

Next, we wish to show that for any fixed time $t\!$, the probability to find a particle in space is equal to one.

The quantity $|\psi ({\textbf {r}},t)|^{2}$ can be interpreted as probability density. To show that this is true, two conditions must be met. First, the probability amplitude must be positive semi-definite (equal to or greater than zero). This condition is trivial because $|\psi ({\textbf {r}},t)|^{2}\!$ is always a positive function. Second, the probability amplitude must be conserved. This condition can be shown by proving that if the wavefunction is normalized at some time $t_{0}\!$ then it must be normalized for any time $t\!$:

$\int _{-\infty }^{\infty }d^{3}r|\psi ({\textbf {r}},t_{0})|^{2}=1\Rightarrow \int d^{3}r|\psi ({\textbf {r}},t)|^{2}=1\;\;\forall t$

The solution to the Schrödinger equation conserves probability, i.e. the probability to find the particle somewhere in the space does not change with time. To see that it does, consider

$i\hbar {\frac {\partial }{\partial t}}\psi ({\textbf {r}},t)=\left[-{\frac {\hbar ^{2}}{2m}}\nabla ^{2}+V({\textbf {r}})\right]\psi ({\textbf {r}},t)$

Now multiply both sides by the complex conjugate of $\psi ({\textbf {r}},t)\!$:

$i\hbar \psi ^{*}({\textbf {r}},t){\frac {\partial }{\partial t}}\psi ({\textbf {r}},t)=\psi ^{*}({\textbf {r}},t)\left[-{\frac {\hbar ^{2}}{2m}}\nabla ^{2}+V({\textbf {r}})\right]\psi ({\textbf {r}},t)$

Now, take the complex conjugate of this entire expression:

$-i\hbar \psi ({\textbf {r}},t){\frac {\partial }{\partial t}}\psi ^{*}({\textbf {r}},t)=\psi ({\textbf {r}},t)\left[-{\frac {\hbar ^{2}}{2m}}\nabla ^{2}+V({\textbf {r}})\right]\psi ^{*}({\textbf {r}},t)$

and taking the difference of the above equations, we finally find

${\frac {\partial }{\partial t}}\left(\psi ^{*}({\textbf {r}},t)\psi ({\textbf {r}},t)\right)+{\frac {\hbar }{2im}}\nabla \cdot \left[\psi ^{*}({\textbf {r}},t)\nabla \psi ({\textbf {r}},t)-(\nabla \psi ^{*}({\textbf {r}},t))\psi ({\textbf {r}},t)\right]=0$

Note that this is in the form of a continuity equation

${\frac {\partial }{\partial t}}\rho ({\textbf {r}},t)+\nabla \cdot {\textbf {j}}({\textbf {r}},t)=0$

where

$\rho ({\textbf {r}},t)=\psi ^{*}({\textbf {r}},t)\psi ({\textbf {r}},t)\!$

is the probability density, and

${\textbf {j}}({\textbf {r}},t)={\frac {\hbar }{2im}}\left[\psi ^{*}({\textbf {r}},t)\nabla \psi ({\textbf {r}},t)-(\nabla \psi ^{*}({\textbf {r}},t))\psi ({\textbf {r}},t)\right]$

is the probability current.

Once we know that the densities and currents constructed from the solution of the Schrödinger equation satisfy the continuity equation, it is easy to show that the probability is conserved. To see that note:

${\frac {\partial }{\partial t}}\int d^{3}r|\psi ({\textbf {r}},t)|^{2}=-\int d^{3}r(\nabla \cdot {\textbf {j}})=-\oint d{\textbf {A}}\cdot {\textbf {j}}=0$

where we used the divergence theorem which relates the volume integrals to surface integrals of a vector field. Since the wavefunction is assumed to vanish outside of the boundary, the current vanishes as well. So, this proved the probability of finding the particle in the whole space is independent of time.