# Help with Quantum Mechanics and Continuity Equation

xboy
no, the fourth term, not the third one.

TFM
$$hbar\left(-\rho \frac{\partial \phi}{\partial t} + i\frac{1}{2}\frac{\partial \rho}{\partial t}\right) = \frac{\hbar^2}{2m} \left(\rho(-\phi'^2 + i\phi'') + \frac{i\phi'\rho'}{2} + \left( \frac{\rho'^22\rho - \rho\rho'^2}{4\rho} \right) + \frac{1}{2} i\phi'\rho'\right) + V\rho$$

?

xboy
perfectly alright !

TFM
Okay so:

$$\hbar\left(-\rho \frac{\partial \phi}{\partial t} + i\frac{1}{2}\frac{\partial \rho}{\partial t}\right) = \frac{\hbar^2}{2m} \left(\rho(-\phi'^2 + i\phi'') + \frac{i\phi'\rho'}{2} + \left( \frac{\rho'^22\rho - \rho\rho'^2}{4\rho} \right) + \frac{1}{2} i\phi'\rho'\right) + V\rho$$

Okay, so what should I do now?

xboy
What does the question ask you to do?

TFM
Well, it asks you to split the function up into imaginary and real parts. Have we sorted this out enough to do this now?

xboy
Yup. Go ahead and split it.

TFM
Okay, so:

$$\hbar\left(-\rho \frac{\partial \phi}{\partial t} + i\frac{1}{2}\frac{\partial \rho}{\partial t}\right) = \frac{\hbar^2}{2m} \left(\rho(-\phi'^2 + i\phi'') + \frac{i\phi'\rho'}{2} + \left( \frac{\rho'^22\rho - \rho\rho'^2}{4\rho} \right) + \frac{1}{2} i\phi'\rho'\right) + V\rho$$

Multiply out brackets:

$$-\hbar\rho \frac{\partial \phi}{\partial t} + \hbar i\frac{1}{2}\frac{\partial \rho}{\partial t} = \frac{\hbar^2}{2m}\rho(-\phi'^2 + i\phi'') + \frac{\hbar^2}{2m}\frac{i\phi'\rho'}{2} + \frac{\hbar^2}{2m}\frac{\rho'^22\rho - \rho\rho'^2}{4\rho} + \frac{\hbar^2}{2m}\frac{1}{2} i\phi'\rho' + V\rho$$

So should I just out the parts with an i on one side, and the real parts on the other side of the equals sign (ie i f(x) = g(x))?

xboy
No. If you have a complex equation of the form

f+ ig = p +is where f,g,p,s are real, it follows that f=p and g=s. Here you have to equate the coefficients of the number i on both sides.

TFM
Okay, so:

$$-\hbar\rho \frac{\partial \phi}{\partial t} + \hbar i\frac{1}{2}\frac{\partial \rho}{\partial t} = \frac{\hbar^2}{2m}\rho(-\phi'^2 + i\phi'') + \frac{\hbar^2}{2m}\frac{i\phi'\rho'}{2} + \frac{\hbar^2}{2m}\frac{\rho'^22\rho - \rho\rho'^2}{4\rho} + \frac{\hbar^2}{2m}\frac{1}{2} i\phi'\rho' + V\rho$$

So if we start on the left side:

$$-\hbar\rho \frac{\partial \phi}{\partial t} + \hbar i\frac{1}{2}\frac{\partial \rho}{\partial t}$$

We have:

$$-\hbar\rho \frac{\partial \phi}{\partial t} + i\left(\frac{\hbar}{2}\frac{\partial \rho}{\partial t}\right)$$

Does this look okay?

xboy
Yes. go on.

TFM
Okay, now for the right side:

$$\frac{\hbar^2}{2m}\rho(-\phi'^2 + i\phi'') + \frac{\hbar^2}{2m}\frac{i\phi'\rho'}{2} + \frac{\hbar^2}{2m}\frac{\rho'^22\rho - \rho\rho'^2}{4\rho} + \frac{\hbar^2}{2m}\frac{1}{2} i\phi'\rho' + V\rho$$

$$-\frac{\hbar^2\rho\phi'^2 }{2m} + \frac{\hbar^2\rho i\phi''}{2m} + \frac{i\phi'\rho'\hbar^2}{4m} + \frac{\hbar^2\rho'^22\rho - \hbar^2\rho\rho'^2}{8m\rho} + \frac{\hbar^2 i\phi'\rho' }{4m}+ V\rho$$

Thus:

$$-\frac{\hbar^2\rho\phi'^2 }{2m} +\frac{\hbar^2\rho'^22\rho - \hbar^2\rho\rho'^2}{8m\rho}+ V\rho + \frac{\hbar^2\rho i\phi''}{2m} + \frac{i\phi'\rho'\hbar^2}{4m} + \frac{\hbar^2 i\phi'\rho' }{4m}$$

$$-\frac{\hbar^2\rho\phi'^2 }{2m} +\frac{\hbar^2\rho'^22\rho - \hbar^2\rho\rho'^2}{8m\rho}+ V\rho + i\left(\frac{\hbar^2\rho \phi''}{2m} + \frac{\phi'\rho'\hbar^2}{4m} + \frac{\hbar^2 \phi'\rho' }{4m} \right)$$

Does this look okay?

xboy
Absolutely okay. Go on, you're almost there.

TFM
So now I have to take the imaginary part. Would that be:

$$i\left(\frac{\hbar}{2}\frac{\partial \rho}{\partial t}\right) = i\left(\frac{\hbar^2\rho \phi''}{2m} + \frac{\phi''\rho''\hbar^2}{4m} + \frac{\hbar^2 \phi''\rho''}{4m} \right)$$

xboy
Quite correct.

TFM
So now I need to compare this:

$$i\left(\frac{\hbar}{2}\frac{\partial \rho}{\partial t}\right) = i\left(\frac{\hbar^2\rho \phi''}{2m} + \frac{\phi'\rho''\hbar^2}{4m} + \frac{\hbar^2 \phi''\rho''}{4m} \right)$$

to:

$$-\frac{\partial}{\partial t}\rho(x,t) = \frac{\hbar}{m}\rho(x,t)\phi''(x,t) + \frac{\hbar}{m}\rho'(x,t)\phi'(x,t)$$

Well is we

$$\frac{\partial \rho}{\partial t} = \frac{\hbar\rho \phi''}{m} + \frac{\phi'\rho''\hbar}{2m} + \frac{\hbar \phi''\rho''}{2m}$$

They don't seem to similar...?

xboy
Oh. Your previous expression was actually incorrect. The last 2 terms in the rhs should have contained single derivatives of both rho and phi, and not double derivativs (compare with the expression prior to that).

TFM
$$\frac{\partial \rho}{\partial t} = \frac{\hbar\rho \phi''}{m} + \frac{\phi'\rho'\hbar}{2m} + \frac{\hbar \phi'\rho'}{2m}$$

so,

$$\frac{\partial \rho}{\partial t} = \frac{\hbar\rho \phi''}{m} + \frac{\phi'\rho'\hbar}{m}$$

Now this does look more similar...