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...