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{{Quantum Mechanics A}}
The quantity <math>|\Psi(\textbf{r},t)|^2</math> can be interpreted as probability density. In order for us to do so, 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 <math>|\Psi(\textbf{r},t)|^2 \!</math> is always a non-negative function. Second, the probability density, integrated over all space, must be equal to one:

:<math>\int_{-\infty}^{\infty}d^3\textbf{r}\,|\Psi(\textbf{r},t)|^2=1</math>

We will show that, if this relation is satisfied for a specific time, then it is satisfied for all times shortly.

Because of the fact that we may interpret <math>|\Psi(\textbf{r},t)|^2</math> as a probability density, we may calculate expectation values of observables, such as position and momentum, in terms of it.  In general, the expectation value of an observable <math>Q(\textbf{r},\textbf{p};t)</math> is given by

:<math>\langle Q(\textbf{r},\textbf{p};t)\rangle=\int_{-\infty}^{\infty}d^3\textbf{r}\,\Psi^{\ast}(\textbf{r},t)Q\left (\textbf{r},-i\hbar\nabla;t\right )\Psi(\textbf{r},t).</math>

In particular, the expectation value of a position coordinate <math>x_i</math> is

:<math>\langle x_i\rangle=\int_{-\infty}^{\infty}d^3\textbf{r}\,x_i|\Psi(\textbf{r},t)|^2</math>

and that for a component of momentum <math>p_i</math> is

:<math>\langle p_i\rangle=\int_{-\infty}^{\infty}d^3\textbf{r}\,\Psi^{\ast}(\textbf{r},t)\left (-i\hbar\frac{\partial}{\partial x_i}\right )\Psi(\textbf{r},t).</math>

== Conservation of Probability ==

We will now show that 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 

<math>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)</math> 

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

<math>i\hbar\Psi^{\ast}(\textbf{r},t)\frac{\partial}{\partial t}\Psi(\textbf{r},t)=\Psi^{\ast}(\textbf{r},t)\left[-\frac{\hbar^2}{2m}\nabla^2+V(\textbf{r})\right]\Psi(\textbf{r},t)</math>

Now, take the complex conjugate of this entire expression:

<math>-i\hbar\Psi(\textbf{r},t)\frac{\partial}{\partial t}\Psi^{\ast}(\textbf{r},t)=\Psi(\textbf{r},t)\left[-\frac{\hbar^2}{2m}\nabla^2+V(\textbf{r})\right]\Psi^{\ast}(\textbf{r},t)</math>

Taking the difference of the above equations, we finally find

<math>\frac{\partial}{\partial t} \left( \Psi^{\ast}(\textbf{r},t)\Psi(\textbf{r},t)\right) + \frac{\hbar}{2im} \nabla \cdot \left[\Psi^{\ast}(\textbf{r},t)\nabla \Psi(\textbf{r},t)-(\nabla\Psi^{\ast}(\textbf{r},t)) \Psi(\textbf{r},t)\right]=0</math>
 
Note that this is in the form of a continuity equation,

<math>\frac{\partial}{\partial t} \rho(\textbf{r},t) + \nabla \cdot \textbf{j}(\textbf{r},t)=0,</math>
 
where 

<math>\rho(\textbf{r},t)=\Psi^{\ast}(\textbf{r},t)\Psi(\textbf{r},t)\!</math> 

is the probability density, and  

:<math>\textbf{j}(\textbf{r},t)=-\frac{i\hbar}{2m}\left[\Psi^{\ast}(\textbf{r},t)\nabla \Psi(\textbf{r},t)-(\nabla\Psi^{\ast}(\textbf{r},t)) \Psi(\textbf{r},t)\right]</math> 

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 this, note that

:<math>\frac{d}{dt}\int d^3\textbf{r} |\Psi(\textbf{r},t)|^2=-\int d^3\textbf{r}(\nabla\cdot \textbf{j})=-\oint d\textbf{A} \cdot \textbf{j} =0.</math> 

Here, we used the fact that the wave function is assumed to vanish outside of the boundary, and thus the current vanishes as well.  Therefore, we see that the normalization of the wave function does not change over time, so that we only need to normalize it at one instant in time, as asserted earlier.

==Problems==

'''(1)''' Consider a particle moving in a potential field <math>V(\textbf{r}).</math>

'''(a)''' Prove that the average energy is <math>\langle E\rangle=\int W\,d^3\textbf{r}=\int\left (\frac{\hbar^2}{2m}\nabla\Psi^{\ast}\cdot\nabla\Psi+\Psi^{\ast}V\Psi\right )\,d^3\textbf{r},</math> where <math>W</math> is energy density.

'''(b)''' Prove the energy conservation equation, <math>\frac{\partial W}{\partial t}+\nabla \cdot \textbf{S}=0,</math> where <math>\textbf{S}=-\frac{\hbar^2}{2m}\left (\frac{\partial\Psi^{\ast}}{\partial t}\nabla\Psi + \frac{\partial\Psi}{\partial t}\nabla\Psi^{\ast}\right )</math> is the energy flux density.

[[Phy5645/Energy_conservation|Solution]]


'''(2)''' Assume that the Hamiltonian for a system of <math>N</math> particles is <math>\hat{H}=-\sum_{i=1}^{N}\frac{\hbar}{2m}\nabla_{i}^{2}+\sum_{i=1}^{N}\rho_{ij}(|\textbf{r}_{i}-\textbf{r}_{j}|)</math>, and <math>\Psi(\textbf{r}_{1},\textbf{r}_{2},\ldots,\textbf{r}_{N};t)</math> is the wave fuction.  Defining

<math>\rho(\textbf{r},t)=\sum\rho_{i}(\textbf{r},t),</math>

<math>\textbf{j}(\textbf{r},t)=\sum\textbf{j}_{i}(\textbf{r},t),</math>

<math>\rho_{1}(\textbf{r}_{1},t)=\int\cdots\int d^{3}\textbf{r}_{2}\,d^{3}\textbf{r}_{3}\,\cdots\,d^{3}\textbf{r}_{N}\,\Psi^{\star}\Psi,</math>

<math>\textbf{j}_{1}(\textbf{r}_{1},t)=-\frac{i\hbar}{2m}\int\cdots\int d^{3}\textbf{r}_{2}\,d^{3}\textbf{r}_{3}\,\cdots\,d^{3}\textbf{r}_{N}\,(\Psi^{\star}\nabla_{1}\Psi-\Psi\nabla_{1}\Psi^{\star}),</math>

and similarly for the other <math>\rho_i</math> and <math>\textbf{j}_i</math>, prove the following relation:

<math>\frac{\partial\rho}{\partial t}+\nabla\cdot\textbf{j}=0</math>

[[Phy5645/schrodingerequationhomework2|Solution]]