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I have a wavefunction $\psi = \pi^{-1\over 4}(1+it)^{-1\over 2} \exp{({-x^2\over 2(1+it)})}$

I want to show that it satisfies  $\partial_t P +\partial_x j =0$  (*)

I calculated $P=\pi^{-1\over 2}(1+t^2)^{-1\over 2} \exp{({-x^2\over (1+t^2)})}$and  $j=ix\pi^{-1\over 2}(1+t^2)^{-3\over 2} \exp{({-x^2\over (1+t^2)})}$

This gives $\partial_t P = -t\pi^{-1\over 2}(1+t^2)^{-5\over 2} (t^2-2x^2+1)\exp{({-x^2\over (1+t^2)})}$ and $\partial_x j = i\pi^{-1\over 2}(1+t^2)^{-5\over 2} (t^2-2x^2+1)\exp{({-x^2\over (1+t^2)})}$ 

But how do they satisfy (*)? Perhaps I have got something wrong? 

Thanks.

1 Answers 1

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First of all, the current is

$j= \frac{i}{2}(\psi\partial_x\psi^*-\psi^*\partial_x\psi) \; ,$

implying that for your choice of wave function

$j=t x \pi^{-1/2}(1+t^2)^{-3/2}\exp\left(-\frac{x^2}{1+t^2}\right) \; .$

Try computing $\partial_x j$ again, it should work out fine now.

EDIT:

So,

$\partial_x\psi= \pi^{-1/4}(1-it)^{-1/2}\left(-\frac{x}{1-it}\right)\exp\left(-\frac{x^2}{2(1-it)}\right) \; ,$

and

$\partial_x\psi^*= \pi^{-1/4}(1+it)^{-1/2}\left(-\frac{x}{1+it}\right)\exp\left(-\frac{x^2}{2(1+it)}\right) \; .$

Multiplying by $\psi^*$ and $\psi$ respectively, one gets

$\psi^*\partial_x\psi= \pi^{-1/2}(1+t^2)^{-1/2}\left(-\frac{x}{1+it}\right)\exp\left(-\frac{x^2}{(1+t^2)}\right) \; ,$

and

$\psi\partial_x\psi^*= \pi^{-1/2}(1+t^2)^{-1/2}\left(-\frac{x}{1-it}\right)\exp\left(-\frac{x^2}{(1+t^2)}\right) \; .$

Finally, subtract both and multiply by $i/2$ to get the current $j$.

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    Ahhhh! Thanks! I know what went wrong now, I have been stupidly adding the complex conjugates, despite knowing that I should be subtracting them. Thanks again!2011-10-17