Probability of Finding a Particle | Quantum Science Philippines

## Prove that the total probability of finding a particle is conserved.

The norm of a state vector is given by:  $\frac{d}{dt} \langle \psi (t) | \psi (t) \rangle$.

In order to prove that the probability of finding a particle is conserved, we must obtain that

$\frac{d}{dt} \langle \psi (t) | \psi (t) \rangle = constant$

Now, we recall that from the 6th Postulate of Quantum Mechanics, the time evolution of the state vector $| \psi (t) \rangle$ is governed by the Schrodinger Equation as,

$i\hbar \frac{d}{dt} | \psi (t) \rangle = H(t)|\psi (t) \rangle$                   [1]

$\frac{d}{dt} | \psi (t) \rangle = \frac{1}{i \hbar} H(t)|\psi (t) \rangle$               [2]

Taking the conjugate,

$\frac{d}{dt} \langle \psi (t)| = -\frac{1}{i \hbar} \langle \psi (t)|H^\dagger (t)$                [3]

but since $H(t)$ by definition is Hermitian, that is, $H^\dagger (t) = H(t)$, then

$\frac{d}{dt} \langle \psi (t) | = - \frac{1}{i \hbar} \langle \psi (t) | H (t)$                 [4]

Now, taking the scalar product of equations [4] and [2],

$\frac{d}{dt} \langle \psi (t) | \psi (t) \rangle = \Big[ \frac{d}{dt} \langle \psi (t)| \Big] | \psi (t) \rangle + \langle \psi (t)| \Big[ \frac{d}{dt} | \psi (t) \rangle \Big]$                 [5]

Substituting equations [2] and [4],

$\frac{d}{dt} \langle \psi (t)| \psi (t) \rangle = - \frac{1}{i \hbar} \langle \psi (t)| H(t)| \psi (t) \rangle + \frac{1}{i \hbar} \langle \psi (t)| H(t)| \psi (t) \rangle$    [6]

$= \frac{1}{i \hbar} \langle \psi (t)| H(t) - H(t) | \psi (t) \rangle$        [7]

$= \frac{1}{i \hbar} \langle \psi (t)| 0 | \psi (t) \rangle$                            [8]

$\frac{d}{dt} \langle \psi (t) | \psi (t) \rangle = 0$                                                                            [9]

Hence, the total probability of finding a particle is conserved , since we obtained that

$\mathbf{\frac{d}{dt} \langle \psi (t) | \psi (t) \rangle} = \mathbf{0}$.    $\Box$