Help with Quantum Mechanics and Continuity Equation

[TFM]

Okay so:

i\hbar\left(\rho i \frac{\partial \phi}{\partial t} + \frac{1}{2}\frac{\partial \rho}{\partial t}\right) = \frac{\hbar^2}{2m} \left(\rho(i\phi&#039;i \phi&#039; + i\phi&#039;&#039;) + \frac{i\phi&#039;\rho&#039;}{2} + \left( \frac{\rho\rho&#039;\rho&#039;2\rho - \rho\rho&#039;^2}{4\rho} \right) + \frac{\rho\rho&#039;}{2} i\phi\rho&#039;\right) + V\rho <br />

Okay so far?

 

[xboy]

On the third term on the rhs you got an extra rho which wasn't there in your previous steps. Also you can multiply the i s in the first term of the rhs to get -1.

 

[TFM]

Okay,

i\hbar\left(\rho i \frac{\partial \phi}{\partial t} + \frac{1}{2}\frac{\partial \rho}{\partial t}\right) = \frac{\hbar^2}{2m} \left(\rho(i\phi&#039;i \phi&#039; + i\phi&#039;&#039;) + \frac{i\phi&#039;\rho&#039;}{2} + \left( \frac{\rho&#039;\rho&#039;2\rho - \rho\rho&#039;^2}{4\rho} \right) + \frac{\rho\rho&#039;}{2} i\phi\rho&#039;\right) + V\rho

and:

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&#039;^2 + i\phi&#039;&#039;) + \frac{i\phi&#039;\rho&#039;}{2} + \left( \frac{\rho&#039;^22\rho - \rho\rho&#039;^2}{4\rho} \right) + \frac{\rho\rho&#039;}{2} i\phi\rho&#039;\right) + V\rho

Does this look better?

 

[xboy]

No, it looks to me that the fourth term on the rhs has one extra rho and one extra rho prime, while the prime on phi has disappeared.

 

[TFM]

Okay:

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&#039;^2 + i\phi&#039;&#039;) + \frac{i\phi&#039;\rho&#039;}{2} + \left( \frac{\rho&#039; 2\rho - \rho&#039;^2}{4\rho} \right) + \frac{\rho\rho&#039;}{2} i\phi&#039;\rho&#039;\right) + V\rho

Better?

 

[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&#039;^2 + i\phi&#039;&#039;) + \frac{i\phi&#039;\rho&#039;}{2} + \left( \frac{\rho&#039;^22\rho - \rho\rho&#039;^2}{4\rho} \right) + \frac{1}{2} i\phi&#039;\rho&#039;\right) + V\rho <br />

?

 

[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&#039;^2 + i\phi&#039;&#039;) + \frac{i\phi&#039;\rho&#039;}{2} + \left( \frac{\rho&#039;^22\rho - \rho\rho&#039;^2}{4\rho} \right) + \frac{1}{2} i\phi&#039;\rho&#039;\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&#039;^2 + i\phi&#039;&#039;) + \frac{i\phi&#039;\rho&#039;}{2} + \left( \frac{\rho&#039;^22\rho - \rho\rho&#039;^2}{4\rho} \right) + \frac{1}{2} i\phi&#039;\rho&#039;\right) + V\rho

Multiply out brackets:

-\hbar\rho \frac{\partial \phi}{\partial t} + \hbar i\frac{1}{2}\frac{\partial \rho}{\partial t} =<br /> \frac{\hbar^2}{2m}\rho(-\phi&#039;^2 + i\phi&#039;&#039;) + \frac{\hbar^2}{2m}\frac{i\phi&#039;\rho&#039;}{2} + \frac{\hbar^2}{2m}\frac{\rho&#039;^22\rho - \rho\rho&#039;^2}{4\rho} + \frac{\hbar^2}{2m}\frac{1}{2} i\phi&#039;\rho&#039; + 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&#039;^2 + i\phi&#039;&#039;) + \frac{\hbar^2}{2m}\frac{i\phi&#039;\rho&#039;}{2} + \frac{\hbar^2}{2m}\frac{\rho&#039;^22\rho - \rho\rho&#039;^2}{4\rho} + \frac{\hbar^2}{2m}\frac{1}{2} i\phi&#039;\rho&#039; + 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&#039;^2 + i\phi&#039;&#039;) + \frac{\hbar^2}{2m}\frac{i\phi&#039;\rho&#039;}{2} + \frac{\hbar^2}{2m}\frac{\rho&#039;^22\rho - \rho\rho&#039;^2}{4\rho} + \frac{\hbar^2}{2m}\frac{1}{2} i\phi&#039;\rho&#039; + V\rho

-\frac{\hbar^2\rho\phi&#039;^2 }{2m} + \frac{\hbar^2\rho i\phi&#039;&#039;}{2m} + \frac{i\phi&#039;\rho&#039;\hbar^2}{4m} + \frac{\hbar^2\rho&#039;^22\rho - \hbar^2\rho\rho&#039;^2}{8m\rho} + \frac{\hbar^2 i\phi&#039;\rho&#039; }{4m}+ V\rho

Thus:

-\frac{\hbar^2\rho\phi&#039;^2 }{2m} +\frac{\hbar^2\rho&#039;^22\rho - \hbar^2\rho\rho&#039;^2}{8m\rho}+ V\rho + \frac{\hbar^2\rho i\phi&#039;&#039;}{2m} + \frac{i\phi&#039;\rho&#039;\hbar^2}{4m} + \frac{\hbar^2 i\phi&#039;\rho&#039; }{4m}


-\frac{\hbar^2\rho\phi&#039;^2 }{2m} +\frac{\hbar^2\rho&#039;^22\rho - \hbar^2\rho\rho&#039;^2}{8m\rho}+ V\rho + i\left(\frac{\hbar^2\rho \phi&#039;&#039;}{2m} + \frac{\phi&#039;\rho&#039;\hbar^2}{4m} + \frac{\hbar^2 \phi&#039;\rho&#039; }{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&#039;&#039;}{2m} + \frac{\phi&#039;&#039;\rho&#039;&#039;\hbar^2}{4m} + \frac{\hbar^2 \phi&#039;&#039;\rho&#039;&#039;}{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&#039;&#039;}{2m} + \frac{\phi&#039;\rho&#039;&#039;\hbar^2}{4m} + \frac{\hbar^2 \phi&#039;&#039;\rho&#039;&#039;}{4m} \right)

to:

-\frac{\partial}{\partial t}\rho(x,t) = \frac{\hbar}{m}\rho(x,t)\phi&#039;&#039;(x,t) + \frac{\hbar}{m}\rho&#039;(x,t)\phi&#039;(x,t)

Well is we

\frac{\partial \rho}{\partial t} = \frac{\hbar\rho \phi&#039;&#039;}{m} + \frac{\phi&#039;\rho&#039;&#039;\hbar}{2m} + \frac{\hbar \phi&#039;&#039;\rho&#039;&#039;}{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&#039;&#039;}{m} + \frac{\phi&#039;\rho&#039;\hbar}{2m} + \frac{\hbar \phi&#039;\rho&#039;}{2m}

so,

\frac{\partial \rho}{\partial t} = \frac{\hbar\rho \phi&#039;&#039;}{m} + \frac{\phi&#039;\rho&#039;\hbar}{m}

Now this does look more similar...