Relation Between the Wave Function and Probability Density: Difference between revisions
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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 | 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}\ | :<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>: | Now multiply both sides by the complex conjugate of <math>\psi(\textbf{r},t) \!</math>: | ||
:<math>i\hbar\ | :<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> <br/> | ||
Now, take the complex conjugate of this entire expression: <br/> | Now, take the complex conjugate of this entire expression: <br/> | ||
:<math>-i\hbar\ | :<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> <br/> | ||
and taking the difference of the above equations, we finally find | and taking the difference of the above equations, we finally find | ||
:<math>\frac{\partial}{\partial t} \left( \ | :<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> | ||
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where | where | ||
:<math>\rho(\textbf{r},t)=\ | :<math>\rho(\textbf{r},t)=\Psi^{\ast}(\textbf{r},t) \Psi(\textbf{r},t)\!</math> | ||
is the probability density, and | is the probability density, and | ||
:<math>\textbf{j}(\textbf{r},t)=\frac{\hbar}{2im}\left[\ | :<math>\textbf{j}(\textbf{r},t)=\frac{\hbar}{2im}\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. | 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. | 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: | To see that note: | ||
:<math>\frac{ | :<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> | ||
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. | 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. Therefore, we see that the normalization of the wave function does not change over time, and therefore we only need to normalize it at one instant in time, as asserted earlier. |
Revision as of 16:35, 8 April 2013
The quantity can be interpreted as probability density. In order for this to be 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 is always a positive function. Second, the probability density, integrated over all space, must be equal to one:
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 as a probability density, we may calculate expectation values of observables, such as position and momentum, in terms of it. These expectation values are
and
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
Now multiply both sides by the complex conjugate of :
Now, take the complex conjugate of this entire expression:
and taking the difference of the above equations, we finally find
Note that this is in the form of a continuity equation
where
is the probability density, and
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:
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. Therefore, we see that the normalization of the wave function does not change over time, and therefore we only need to normalize it at one instant in time, as asserted earlier.