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.


1 Answer 1


Multiply the equation by $\psi^*$:

$$\psi^* \frac{\partial}{\partial t} \psi = -i \psi^* \frac{\partial^2}{\partial x^2} \psi$$

We note that

$$2 \Re{\left [ \psi^* \frac{\partial}{\partial t} \psi \right ]} = \frac{\partial}{\partial t} [\psi^* \psi]$$


$$\Re{\left[\int_0^{\ell} dx \:\psi^* \frac{\partial}{\partial t} \psi\right]} =\frac{1}{2} \frac{\partial}{\partial t} \int_0^{\ell} dx \:[\psi^* \psi] = \Re{ \left [-i \int_0^{\ell} dx \:\psi^* \frac{\partial^2}{\partial x^2} \psi\right ]}$$

We integrate the integral on the RHS by parts:

$$\int_0^{\ell} dx \:\psi^* \frac{\partial^2}{\partial x^2} \psi = \underbrace{\left [ \psi^* \frac{\partial}{\partial x} \psi \right ]_0^{\ell}}_{\text{this}=0} - \int_0^{\ell} dx \: \left | \frac{\partial \psi}{\partial x} \right |^2$$


$$ \frac{\partial}{\partial t} \int_0^{\ell} dx \: |\psi|^2 = 2 \Re{\left [-i \int_0^{\ell} dx \:\left | \frac{\partial \psi}{\partial x} \right |^2 \right ]}$$

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

$$\displaystyle \int_0^l dx\:\mid\psi(t, x)\mid^2 = \int_0^l dx\: \mid\psi(0, x)\mid^2 $$

  • $\begingroup$ Thank you - but I don't understand the first step. Which equation is being multiplied by $\psi^\*$? And what is the meaning of $\psi^\*$? $\endgroup$
    – Lizards
    Apr 15, 2013 at 8:53
  • $\begingroup$ @Lily: the Schroedinger equation you posted. $\psi^*$ is the complex conjugate of $\psi$. $\endgroup$
    – Ron Gordon
    Apr 15, 2013 at 11:13

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