I'm given a complex wave-function $\psi(t, x)$, in one spatial dimension, which satisfies $i \partial_t \psi = \partial_x^2 \psi$, a simplified form of Schrodinger's equation in one spatial dimension. The Dirichlet boundary conditions on the function are $\psi(t, 0) = \psi(t, l) = 0$.
Using the fact that $\mid\psi\mid^2 = \psi \bar{\psi}$ and $-i\bar{\psi_t} = \bar{\psi_{tt}}$ I have shown that $\partial_t \mid\psi^2\mid = \frac{1}{i} \partial_x (\psi_x \bar{\psi} - \bar{\psi_x} \psi)$.
However, I don't understand how to also show:
$\displaystyle \int_0^l \mid\psi(t, x)\mid^2 \mathrm{d}x = \int_0^l \mid\psi(0, x)\mid^2 \mathrm{d}x$
Which would imply that the particle described by the wave-function could not escape its bounding box. I just don't see the connection, and I would appreciate any amount of explanation someone can provide.
Multiply the equation by $\psi^*$:
$$\psi^* \frac{\partial}{\partial t} \psi = -i \psi^* \frac{\partial^2}{\partial x^2} \psi$$
We note that
$$2 \Re{\left [ \psi^* \frac{\partial}{\partial t} \psi \right ]} = \frac{\partial}{\partial t} [\psi^* \psi]$$
Then
$$\Re{\left[\int_0^{\ell} dx \:\psi^* \frac{\partial}{\partial t} \psi\right]} =\frac{1}{2} \frac{\partial}{\partial t} \int_0^{\ell} dx \:[\psi^* \psi] = \Re{ \left [-i \int_0^{\ell} dx \:\psi^* \frac{\partial^2}{\partial x^2} \psi\right ]}$$
We integrate the integral on the RHS by parts:
$$\int_0^{\ell} dx \:\psi^* \frac{\partial^2}{\partial x^2} \psi = \underbrace{\left [ \psi^* \frac{\partial}{\partial x} \psi \right ]_0^{\ell}}_{\text{this}=0} - \int_0^{\ell} dx \: \left | \frac{\partial \psi}{\partial x} \right |^2$$
Therefore
$$ \frac{\partial}{\partial t} \int_0^{\ell} dx \: |\psi|^2 = 2 \Re{\left [-i \int_0^{\ell} dx \:\left | \frac{\partial \psi}{\partial x} \right |^2 \right ]}$$
The RHS is zero because it is the real part of a purely imaginary number. Therefore the left hand side is also zero. Thus
$$\int_0^{\ell} dx \: |\psi|^2$$
is independent of $t$. Therefore
$$\displaystyle \int_0^l dx\:\mid\psi(t, x)\mid^2 = \int_0^l dx\: \mid\psi(0, x)\mid^2 $$