Conservation of probability: Suppose a wavefunction has ${\partial \mathbb P \over \partial t} = -t f(x,t)$ and ${\partial j \over \partial x} = i f(x,t)$. How does it follow that ${\partial \mathbb P \over \partial t} = {-\partial j \over \partial x}$? Thanks.
Tell me more
×
Physics Stack Exchange is a question and answer site for
active researchers, academics and students of physics. It's 100% free, no registration required.
|
|
||||
|
|
Are you looking for a proof? If so, this link (which has some sign errors as pointed out in the comments) proves it as follows (without the sign errors): We start by differentiating the definition of the probability with respect to time only: $$ \frac{\partial P(x,t)}{\partial t} = \frac{\partial}{\partial t}\left (\psi^*(x,t) \psi(x,t)\right) = \left[ \frac{\partial\psi^*}{\partial t}\psi + \psi^*\frac{\partial\psi}{\partial t} \right] (1) $$ We now exploit Schrödinger's Equation and its complex conjugate: $$ -\frac{\hbar^2 }{2m}\frac{\partial^2\psi}{\partial x^2} + V(x)\psi = i \hbar \frac{\partial \psi}{\partial t} $$ $$ -\frac{\hbar^2 }{2m}\frac{\partial^2\psi^*}{\partial x^2} + V(x)\psi^* = -i \hbar \frac{\partial \psi^*}{\partial t} $$ And inject them into (1): $$ \frac{\partial P(x,t)}{\partial t} = \frac 1 {i\hbar}\left[ \frac{\hbar^2 }{2m}\frac{\partial^2\psi^*}{\partial x^2}\psi - V(x)\psi^*\psi -\frac{\hbar^2 }{2m}\frac{\partial^2\psi}{\partial x^2}\psi^* + V(x)\psi\psi^* \right] $$ Which corresponds to: $$ \frac{\partial P(x,t)}{\partial t} = \frac 1 {i\hbar} \frac{\hbar^2}{2m} \left[ \frac{\partial^2\psi^*}{\partial x^2}\psi - \psi^*\frac{\partial^2\psi}{\partial x^2}\right] = \frac{\hbar}{2mi}\frac{\partial}{\partial x} \left[ \frac{\partial\psi^*}{\partial x}\psi - \psi^*\frac{\partial\psi}{\partial x} \right] (2) $$ The probability current is defined as follows: $$ j(x,t) = \frac{\hbar}{2mi} \left[\psi^* \frac{\partial \psi}{\partial x} - \psi \frac{\partial \psi^*}{\partial x} \right] $$ Hence, its differential with respect to the $x$ axis is the following: $$ \frac{\partial j(x,t)}{\partial x} = \frac{\hbar}{2mi} \left[\psi^* \frac{\partial ^2 \psi}{\partial x^2} - \psi \frac{\partial^2 \psi^*}{\partial x^2}\right] = \frac{\hbar}{2mi} \frac{\partial}{\partial x} \left[ \psi^* \frac{\partial \psi}{\partial x} - \psi \frac{\partial \psi^*}{\partial x} \right ] (3) $$ $$ \mbox{(2) and (3)} \Leftrightarrow \frac{\partial P(x,t)}{dt} + \frac{\partial j(x,t)}{\partial x} = 0 $$ |
|||||||||||||
|
