# Help with Quantum Mechanics and Continuity Equation

• TFM
no, the fourth term, not the third one.

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

?

perfectly alright !

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?

What does the question ask you to do?

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

Yup. Go ahead and split it.

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))?

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.

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?

Yes. go on.

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?

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

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

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

Quite correct.

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...?

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).

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

so,

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

Now this does look more similar...