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| {{Quantum Mechanics A}} | | {{Quantum Mechanics A}} |
| The quantity <math>|\Psi(\textbf{r},t)|^2</math> 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 <math>|\Psi(\textbf{r},t)|^2 \!</math> is always a positive function. Second, the probability density, integrated over all space, must be equal to one: | | 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 positive function. Second, the probability density, integrated over all space, must be equal to one: |
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| :<math>\int_{-\infty}^{\infty}d^3\textbf{r}\,|\Psi(\textbf{r},t)|^2=1</math> | | :<math>\int_{-\infty}^{\infty}d^3\textbf{r}\,|\Psi(\textbf{r},t)|^2=1</math> |
Revision as of 15:39, 10 April 2013
The quantity
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
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. In general, the expectation value of an observable
is given by

In particular, the expectation value of a position coordinate
is

and that for a component of momentum
is

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
![{\displaystyle 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)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fe5426d4ebc2fc5a606af2474e02f6334d8a8378)
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
![{\displaystyle {\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}](https://wikimedia.org/api/rest_v1/media/math/render/svg/dd22abfbf400ca945743211b83712ac88118d713)
Note that this is in the form of a continuity equation

where

is the probability density, and
![{\displaystyle {\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]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/94de3113892ff15f0e309ede6ca16871ec91fe0a)
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.