Help with Quantum Mechanics and Continuity Equation

[TFM]

Homework Statement

A Bose-Einstein condensate can be described by a wave function

\psi(x,t) = \sqrt{\rho(x,t)}e^{i\phi(x,t)}

Where the functions:

\phi(x,t) and \rho(x,t)

are real.

a)

What is the probability density

b)

Calculate the probability current density as a function of ρ(x,t)and ϕ(x,t) and their derivatives.

c)

Use the results of (a) and (b) and write down the continuity equation in terms of ρ(x,t)and ϕ(x,t) and their derivatives.

d)

Substitute ψ(x,t)=√(ρ(x,t) ) e^iϕ(x,t) into the Schrödinger equation for a one-dimensional particle moving in a potential V (x). (Watch out when taking the derivatives. You need to apply product and chain rules.) Then multiply the whole equation by ψ^* (x,t) in order to simplify it. The resulting equation can be split into two: one for the real and one for the imaginary part. Take the imaginary part and compare the equation you obtain to the result you got under (c).

Homework Equations

Probability Density:

ρ= ψ* ψ

Current Density:

j(x,t)=\frac{\hbar}{2im}\left(\phi^* \frac{d\phi}{dx}-\phi\frac{d\phi^*}{dx}\right)

Continuity Equation:

\nabla \cdot J = -\frac{\partial}{\partial t}\rho(x,t)

The Attempt at a Solution

I have a lot of workings out, So I have attached them as a word document, if that is okay.

I have found (a) to be:

\rho = \rho(x,t)

Which is the right answer.

For (b), I have:

j(x,t) = \frac{\hbar \rho(x,t)\phi'(x,t)}{m} + \frac{\hbar \rho'(x,t)}{2im}

Now I have inserted these into the continuity equation, but I have got:

\nabla \cdot J = \frac{\hbar}{m}\rho(x,t)\phi''(x,t) + \frac{\hbar}{m}\rho'(x,t)\phi'(x,t) + \frac{\hbar}{2im} \rho''(x,t)

Whereas for the other side:

-\frac{\partial}{\partial t} \rho{x,t} = -\rho'{x,t}

Pleas not that the working out for b is on the first word document, and for the c and d are on the second word document.

Any ideas where I may have gone wrong?

Thanks in advanced,

TFM

 

[xboy]

You made an algebraic error in calculating the probability current density. Check the fifth step in your subtraction ( 4 steps above the last step in page 2 of the first document)

 

[TFM]

Okay so I assume that these are the error is located within these three lines:

(I am just writing h, but it represents hbar)

j(x,t) = \frac{h}{2im}\left( \left(\sqrt{\rho}e^{-i\phi}\sqrt{\rho}ie^{i\phi}\phi' + \sqrt{\rho}e^{-i\phi}e^{i\phi}\frac{1}{2\sqrt{\rho}}\rho'(x) \right} - \left(-\sqrt{\rho}e^{i\phi}\sqrt{\rho}ie^{-i\phi}\phi' + \sqrt{\rho}e^{i\phi}e^{-i\phi}\frac{1}{2\sqrt{\rho}}\rho' \right}\right}

j = \frac{h}{2im}\left(\left(i\rho\phi' + \frac{\sqrt{\rho}}{2\sqrt{\rho}}\rho' \right) - \left(-i\rho\phi' + \frac{\sqrt{\rho}}{2\sqrt{\rho}}\rho' \right)\right)

j = \frac{h}{2im}\left( 2\left(i\rho\phi' + \frac{\sqrt{\rho}}{2\sqrt{\rho}}\rho' \right) \right)

I see a problem:

j = \frac{h}{2im}\left(\left(i\rho\phi' + \frac{\sqrt{\rho}}{2\sqrt{\rho}}\rho' \right) + \left(i\rho\phi' - \frac{\sqrt{\rho}}{2\sqrt{\rho}}\rho' \right)\right)

This means that

The last two terms will cancel:

j = \frac{h}{2im}\left(2i\rho\phi' \right)

Does this look better?

 

[xboy]

Yes, this is what you should be getting.

 

[TFM]

Excellent, thanks.

Cancels down to:

j(x,t) = \frac{h2i\rho(x,t)\phi'(x,t)}{2im}

j(x,t) = \frac{\hbar\rho(x,t)\phi'(x,t)}{m}

So:

\rho = \rho(x,t)

j(x,t) = \frac{\hbar\rho(x,t)\phi'(x,t)}{m}

Continuity Equation:

\nabla \cdot J = -\frac{\partial}{\partial t}\rho(x,t)

Does:

-\frac{\partial}{\partial t}\rho(x,t) = \rho'(x,t)

?

 

[xboy]

I thought you are using the ' sign for differentiations w.r.t space. The derivative w.r.t time will of course be something different. You should just write it like this:
<br /> -\frac{\partial}{\partial t}\rho(x,t) =\frac{\hbar}{m}\rho(x,t)\ phisingle-quotesingle-quote(x,t) + \frac{\hbar}{m}\ rhosingle-quote(x,t)\phisingle-quote(x,t)

And then do the rest of the problem.

 

[TFM]

Okay,

-\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)

But how did you get this, since rho was just equal to rho(x,t)?

 

[xboy]

I don't get you. What do you mean?

 

[TFM]

I am not quite sure how you have gone from

\rho = \rho(x,t)

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)

Where has the \psi come from?

 

[xboy]

By equating the two sides of the continuity equation, separately calculated.

I think you have misunderstood the question. You seem to think that (c) asks you to demonstrate that the continuity relation holds. It doesn't ! All it's asking you to do is to write down how the continuity equation will look for a wave function of the form given.

Because you are given arbitrary functions (rho and phi) you can never verify the continuity equation here. After all one side has time derivative, the other space derivatives. Now if you were given some concrete functions like sine, cos stuff like that, you could have verified it.

 

[TFM]

I see,

So it will be best to keep the time derivative as it is, and just sort out the other side, so:

-\frac{\partial}{\partial t}\rho(x,t) = \nabla \cdot j(x,t)

Now since we are only discussing a 1d Particle:

-\frac{\partial}{\partial t}\rho(x,t) = \frac{d}{dx}j(x,t)

so:

\frac{d}{dx}j(x,t)

j(x,t) = \frac{\hbar\rho(x,t)\phi(x,t)}{m}

So the differential, using the product rule, would be:

\frac{d}{dx}j(x,t) = \frac{\hbar}{m}\rho(x,t)\phi&#039;&#039;(x,t) + \frac{\hbar}{m}\rho&#039;(x,t)\phi&#039;(x,t)

Thus:

-\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)

Does this look okay?

 

[xboy]

looks perfectly fine to me.

 

[TFM]

Okay, so is that all I need to so now for that part?

 

[xboy]

Yeah, what else.

 

[TFM]

Okay, it just didn't so hard once I knew where my error's were...

So for the next part,

Substitute \psi(x,t) = \sqrt{\rho(x,t)}e^{i\phi(x,t)} into the Schrödinger equation for a one-dimensional particle moving in a potential V (x). (Watch out when taking the derivatives. You need to apply product and chain rules.) Then multiply the whole equation by ψ* (x,t) in order to simplify it. The resulting equation can be split into two: one for the real and one for the imaginary part. Take the imaginary part and compare the equation you obtain to the result you got under (c).

I just have to stick:

\psi(x,t) = \sqrt{\rho(x,t)}e^{i\phi(x,t)}

Into Schrödinger's Equation,

i\hbar \frac{\partial\psi}{\partial t} = -\frac{\hbar^2}{2m}\frac{\partial^2 \psi}{\partial x^2}+V\psi

Thus:

i\hbar \frac{\partial}{\partial t}\sqrt{\rho(x,t)}e^{i\phi(x,t)} = -\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}\sqrt{\rho(x,t)}e^{i\phi(x,t)} + V\sqrt{\rho(x,t)}e^{i\phi(x,t)}

So now I presume I will have to find the first time derivative, and the second spatial derivative, or should I leave the time derivative as it is, especially since part of the question asks:

Take the imaginary part and compare the equation you obtain to the result you got under (c).

 

[TFM]

Okay, I have taken the first spatial different of psi, and I have got:

\sqrt{\rho(x,t)}i\phi&#039;(x,t)e^{i\phi(x,t)} + e^{i\phi(x,t)}\frac{\rho&#039;(x,t)}{2\sqrt{\rho(x,t)}}

But I have two Subjects to be differentiated again, but there is three different x terms, what would be the best way to differentiate these, since the product rule is only useful for two x terms?

 

[xboy]

1. You have to take the time derivatives.

2.The product rule applies for as many x terms as you have. You should work out why this is so, I'll give you a hint: suppose you have a product of 3 x terms : a(x)b(x) c(x). Now a(x)b(x) is also a function of x which we may call f(x). Try to take it from there.

 

[TFM]

OKay, so if we start with the left spartial derivative:

\sqrt{\rho(x,t)}i\phi&#039;(x,t)e^{i\phi(x,t)}

u = \sqrt{\rho(x,t)}

du/dx = \frac{1}{2\sqrt{\rho(x,t)}}\rho&#039;(x,t)

v = i\phi&#039;(x,t)e^{i\phi(x,t)}

Now using the product rule again:

a = i\phi&#039;(x,t)

da/dx = i\phi&#039;&#039;(x,t)

b = e^{i\phi(x,t)}

db/dx = i\phi&#039;(x,t)e^{i\phi(x,t)}

Thus:

dv/dx = i\phi&#039;(x,t) i\phi&#039;(x,t)e^{i\phi(x,t)} + i\phi&#039;&#039;(x,t)e^{i\phi(x,t)}

Thus:

The second spatial derivative for the left =

\sqrt{\rho(x,t)}(i\phi&#039;(x,t) i\phi&#039;(x,t)e^{i\phi(x,t)} + i\phi&#039;&#039;(x,t)e^{i\phi(x,t)}) + i\phi&#039;(x,t)e^{i\phi(x,t)}(\frac{1}{2\sqrt{\rho(x,t)}}\rho&#039;(x,t))

Does this look okay so far?

 

[TFM]

I have now done the right spatial differential, and using product rle, chain rula and quotient rule, I have got:

e^{i\phi(x,t)}\frac{\rho&#039;&#039;(x,t)2\sqrt{\rho(x,t)} - (\rho&#039;(x,t))^2/\sqrt{\rho(x,t)}}{4\rho(x,t)}

Does this look okay?

How would you represent the time derivative of:

\sqrt{\rho(x,t)}e^{i\phi(x,t)}

 

[xboy]

The second spatial derivative : Alright except you missed the rho term in the second term. The right spatial derivative : you seem to have missed a term: the exp (i phi) term hasn't been differentiated.

Regarding the time derivative, you got to use the product rule again. And you have to keep in mind that you are using the ' sigh for space derivatives, so here you have to use the \frac {\partial} {\partial t} sign.

 

[TFM]

Okay, so the time derivative is:

-\frac{\partial(\sqrt{\rho(x,t)}e^{i\phi(x,t)})}{\partial t}

Right then:

Left Differential:

\sqrt{\rho}(i\phi&#039;i\phi&#039;e^{i\phi} + i\phi&#039;&#039;e^{i\phi})+ \frac{i\phi&#039;e^{i\phi}}{2\sqrt{\rho}\rho&#039;}

Right Differential:

\rho&#039;\left( \frac{\rho&#039;&#039;2\sqrt{\rho} - (\rho&#039;)^2/\sqrt{\rho}}{4\rho} \right) + \frac{\rho&#039;\rho&#039;&#039;}{2\sqrt{\rho}}

Does this look better?

 

[xboy]

1. Use the product rule on the time derivative.

2. The left differential looks fine. But you have mislaid the exp(i phi) term on the right differential. exp(i phi) is one function and the rest is another function. You can use product rule to separate the two, and then quotient rule or whatever it is called to evaluate the derivative of the second term.

 

[TFM]

Okay, so I need to take the product rule for:

-\frac{\partial(\sqrt{\rho(x,t)}e^{i\phi(x,t)})}{\partial t}

So:

u = \sqrt{\rho(x,t)}, du/dt = \frac{1}{2\sqrt{\rho}}\frac{\partial \rho}{\partial t}

v = e^{i\phi}, dv/dt = ie^{i\phi} \frac{\partial \phi}{\partial t}

Thus:

-\frac{\partial(\sqrt{\rho(x,t)}e^{i\phi(x,t)})}{\partial t} = \sqrt{\rho(x,t)} ie^{i\phi} \frac{\partial \phi}{\partial t} + e^{i\phi}\frac{1}{2\sqrt{\rho}}\frac{\partial \rho}{\partial t}

Does this look okay?

 

[xboy]

perfectly okay.

 

[TFM]

Okay, so for that right derivative:

u = e^{i\phi}, du/dx = i\phi&#039;e^{i\phi}

v = \frac{\rho&#039;}{2\sqrt{\rho}}

dv/dx = \frac{\rho&#039;&#039;2\sqrt{\rho} - \frac{\rho&#039;^2}{\sqrt{\rho}}}{4\rho}

Thus:

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

Does this look better?

 

[xboy]

Yeah, this looks alrighty.

 

[TFM]

Okay, so now:

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

So now I have to multiply everything by phi*:

\sqrt{\rho}e^{-i\phi}

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

 

[xboy]

The third term in the rhs of the las t equation should have the derivative of rho above rather than below.

 

[TFM]

So should have been:

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

Better?

 

[xboy]

Yes, now go ahead and finish it off!