# Homework Help: Quantum particle in a magnetic field

1. May 22, 2014

### carllacan

1. The problem statement, all variables and given/known data
A particle with electrical charge $q$ and mass $m$ is in a electromagnetic field described by $\phi (\vec{r}, t)$ and $A(\vec{r}, t)$. Its Hamiltonian is as follows:
$H = \frac{1}{2m} \left ( \frac{\hbar}{i}\vec{\nabla}-\frac{q}{c} \vec{A} (\vec{r}, t) \right ) ^2 +q\phi (\vec{r}, t)$

The conservation of charge guarantees the continuity equation is fulfilled:
$\frac{\partial}{\partial t} \rho (\vec{r},t)+\vec{\nabla}·\vec{j}(\vec{r},t) = 0$,
where $\rho = q\left|\Psi(x)\right|^2$ is the charge density.

Find the current density $\vec{j}(\vec{r},t)$

2. Relevant equations

The Hamiltonian of a particle in an electromagnetic field.
$H = \frac{1}{2m} \left ( \frac{\hbar}{i}\vec{\nabla}-\frac{q}{c} \vec{A} (\vec{r}, t) \right ) ^2 +q\phi (\vec{r}, t)$

The continuity equation
$\frac{\partial}{\partial t} \rho (\vec{r},t)+\vec{\nabla}·\vec{j}(\vec{r},t) = 0$

3. The attempt at a solution
I've tried stating the time-dependent Shcrodinger equation $\hat{H}\Psi (x) = i\hbar\frac{\partial \Psi(x)}{\partial t}$ and solve for the time-derivative of the wavefunction, which gives:
$\frac{\partial \Psi}{\partial t} = \left [ \frac{\vec{\nabla}^2}{2m}+\frac{q}{\hbar imc}\vec{A}\vec{\nabla}-\frac{1}{2\hbar^2 m} \left( \frac{q}{c} \vec{A}\right )^2 -\frac{q}{2 \hbar m}\phi\right]\Psi$

Then

$\frac{\partial\rho}{\partial t} = q2|\Psi(x)|\frac{\partial |\Psi(x)|}{\partial t} = \left [ \frac{q\vec{\nabla}^2}{m}+\frac{2q^2}{\hbar imc}\vec{A}\vec{\nabla}-\frac{q}{\hbar^2 m} \left( \frac{q}{c} \vec{A}\right )^2 -\frac{q^2}{ \hbar m}\phi\right]\Psi^2$

And then, from the continuity equation we get
$\vec{\nabla}·\vec{j}(\vec{r},t) = \frac{\partial}{\partial t} \rho (\vec{r},t) = \left [ \frac{q\vec{\nabla}^2}{m}+\frac{2q^2}{\hbar imc}\vec{A}\vec{\nabla}-\frac{q}{\hbar^2 m} \left( \frac{q}{c} \vec{A}\right )^2 -\frac{q^2}{ \hbar m}\phi\right]\Psi^2$

But I'm stuck there. How do I "remove" the nabla operators?

Last edited: May 22, 2014
2. May 22, 2014

### carllacan

Question edited with more information.

3. May 22, 2014

### ChrisVer

You must be careful at your steps and how you deal with complex functions and operators.
Problematic points in your approach
No1: your derivative of $\rho$ is not a useful thing... it's better to try to write $|\psi|^{2}= \psi^{\dagger} \psi$ and then use the derivative on each... $(\frac{\partial \psi}{\partial t}) \psi^{\dagger} + \psi \frac{\partial \psi^{\dagger}}{\partial t}$
No2: you had $|\psi|$ on the left and then you moved it on the right, passing it through operators like nabla to write $\psi^{2}$
No3: You don't know the derivative of $|\psi|$... as you've written it, it must be a real number, yet you have real or complex operators $\frac{q^{2}}{hm} \phi$ or $\frac{2q^{2}}{himc} A∇$ acting on a complex wavefunction...

4. May 23, 2014

### carllacan

Yeah, that didn't feel right at all. Should I have left the wavefunction at the left of the operators after expanding the squared term?

I don't get what's the problem with my derivative. Can't I use the chain rule?

5. May 23, 2014

### ChrisVer

your time derivative gives you quantities that are not easy to work on... Schroedinger's eq. gives you the time evolution of the wavefunction, not its module. Both yours and mine use the chain rule, but yours is not useful- you cannot work easily on it, if you can at all... Doing the otherway, you can always take the conjugate of the Schr eq and insert the expression for $\frac{\partial \psi^{*}}{\partial t}$
So even if you leave your wf at the left, things are not right because you put wrong expression for the time derivative.

6. May 23, 2014

### carllacan

Oh, right, what I did looks stupid now. Thank you!

Now: how would the nabla operator work on the wavefunction? Since $\Psi(x)$ is an scalar I'm guessing:
$\vec{\nabla}\Psi(x) = \left (\frac{\partial \Psi(x)}{\partial x}, \frac{\partial \Psi(x)}{\partial y}, \frac{\partial \Psi(x)}{\partial z} \right ) = \frac{\partial \Psi(x)}{\partial x} · \vec{\hat{x}}$, is that right?

And then $\vec{\nabla}^2\Psi(x) = \frac{\partial ^2\Psi(x)}{\partial x^2}$

Last edited: May 23, 2014
7. May 23, 2014

### ChrisVer

No... The $x$ in the wavefunction doesn't really mean the $x$ variable alone- that would be true only if you had 1D problem, and then you nabla would just be the partial of x alone. The $x$ in the wavefunction is a vector, that means:
$\psi(x)= \psi(\vec{x})=\psi(x,y,z)$

why do you want to act with the nabla on the wavefunction in the first place?

8. May 23, 2014

### carllacan

I don't really need that for this problem, I just wanted to know how would that work. Maybe I shouldn't have put it here, sorry.

Using what you said I've managed to find a very simple solution. I'd appreciate if anyone gave a look at it:
$i\hbar \frac{\partial \psi}{\partial t}= \hat{H} \psi \rightarrow \frac{\partial \psi}{\partial t} = \frac{1}{i\hbar} \hat{H} \psi =\frac{1}{2mi\hbar} \left ( \frac{\hbar}{i}\vec{\nabla}-\frac{q}{c} \vec{A} (\vec{r}, t) \right ) ^2 \psi+ \frac{q\phi}{i\hbar} \psi$
Expanding the square:
$\frac{\partial \psi}{\partial t} = \frac{1}{2i\hbar m} \left( -\hbar ^2\vec{\nabla}^2 - \frac{\hbar q}{ic}\vec{\nabla}\vec{A} +\left ( \frac{q}{c} \vec{A}\right )^2\right )\psi + \frac{q\phi}{i\hbar} \psi$
Reorder:
$\frac{\partial \psi}{\partial t} = \left ( \frac{i\hbar}{2m} \vec{\nabla}^2 + \frac{q}{2mc} \vec{\nabla}\vec{A} - \frac{i}{2m\hbar} \left( \frac{q}{c} \vec{A}\right)^2 - \frac{iq\phi}{\hbar} \right )\psi$
And therefore:
$\frac{\partial \psi ^\dagger}{\partial t} = \left ( -\frac{i\hbar}{2m} \vec{\nabla}^2 + \frac{q}{2mc} \vec{\nabla}\vec{A} + \frac{i}{2m\hbar} \left( \frac{q}{c} \vec{A}\right)^2 + \frac{iq\phi}{\hbar} \right )\psi$
Now:
$\frac{\partial \psi}{\partial t}\psi ^\dagger = \left ( \frac{i\hbar}{2m} \vec{\nabla}^2 + \frac{q}{2mc} \vec{\nabla}\vec{A} - \frac{i}{2m\hbar} \left( \frac{q}{c} \vec{A}\right)^2 - \frac{iq\phi}{\hbar} \right )|\psi|^2$ and similarly for $\psi\frac{\partial \psi^\dagger}{\partial t}$
And from there is easy:
$\frac{\partial \rho}{\partial t} = q\frac{\partial }{\partial t}|\psi|^2 = q \frac{\partial}{\partial t} \psi ^\dagger \psi = q \left(\frac{\partial \psi^\dagger}{\partial t} \psi + \psi ^\dagger \frac{\partial \psi}{\partial t} \right) = \left (\frac{q^2}{mc}\vec{\nabla}\vec{A}\right) |\psi|^2$ (because most terms in the parentheses cancel each other).
And then from the continuity equation I finally obtain for $\vec{\nabla}\vec{j}$:
$\vec{\nabla}\vec{j} = -\frac{\partial\rho}{\partial t} = \frac{q^2}{mc}\vec{\nabla}\vec{A} |\psi|^2$

Now my question is: can I just take those nablas out?

9. May 23, 2014

### ChrisVer

You should be more careful when you expand the square of operators....
$∇A \ne A∇$
For this kind of terms you need to do something like:
$∇(A f)= A∇f + (∇A)f$
and then you can use the first term $A∇$ with the same term coming from the square...eg:
$(c ∇+ d A)(c ∇+dA)= c^{2} ∇^{2} + d^{2}A^{2} + cd A∇ + cd ∇A$
in the 3rd term, nabla acts on the wavefunction alone, in the 4th term nabla acts on both A and the wavefunction:
$(c ∇+ d A)^{2}= c^{2} ∇^{2} + d^{2}A^{2} + cd A∇+ cd A∇+ cd (∇A)$
$(c ∇+ d A)^{2}= c^{2} ∇^{2} + d^{2}A^{2} + 2cd A∇+ cd (∇A)$

Also the derivative of the conjugate, you should also take the complex for $\psi$ at the RHS.
Also avoid writing things like $(\hat{O} \psi) \psi^{*}=\hat{O} |\psi|^{2}$
The reason is that in the first case the operator acts on $\psi$ and you suddenly make it act on $|\psi|^{2}$.
What you should do is try to move nablas around to create a needed expression (as the one I give in the end of this post)

In order to move nablas out, you have to do what I did above for the square (work with $∇(AB)=(∇A)B + A(∇B)$)
and don't be in haste to reach a result, think carefully for each point.

Afterwards, if you have an equation:
$∇J= ∇(Something)$
you can deduce that
$J= Something$

Last edited: May 23, 2014
10. May 23, 2014

### ChrisVer

editted^

11. May 25, 2014

### carllacan

Would this be the correct Hamiltonian with the expanded square?
$\hat{H}=\frac{1}{2m}\left(-\hbar^2\vec{\nabla}^2 -\frac{\hbar q}{ic}\left ( \vec{\nabla}·\vec{A} + \vec{A}·\vec{\nabla}\right ) + \left(\frac{q}{c}\vec{A}\right)^2 \right) + q\phi$

Last edited: May 25, 2014
12. Jun 13, 2014

### carllacan

After solving a couple of formalism problems I had I´ve come to this:
$\hat{H}\psi = -\frac{\hbar^2}{2m}\nabla^2\psi-\frac{\hbar q}{2imc}(\vec{\nabla}\cdot\vec{A})\psi+\frac{\hbar q}{imc}\vec{A}\cdot(\vec{\nabla}\psi)+\frac{q^2\vec{A^2}}{2mc^2}\psi+ \frac{q \phi}{2m}\psi$

Can you check if its right?

13. Jun 27, 2014

### carllacan

I understand that in this step the $cd ∇A$ term at the end has been developed into $cd A∇+ cd (∇A)$.

But isn´t $c^{2} ∇^{2} = c^{2}∇∇$ a particular case of $cd ∇A$ with $d = c$ and $A = ∇$?

Why doesn´t the same thing happen with $c^{2} ∇^{2}$? That is, why don´t we have $c^{2} ∇^{2} =c^{2} ∇∇+ c^{2} (∇∇)$?

14. Jun 27, 2014

### carllacan

I've done it again from scratch. Can you tell me if I got it right this time before I go on?
$-\frac{\hbar}{i}\frac{\partial \Psi}{\partial t} = \hat{H}\Psi$
$-\frac{\hbar}{i}\frac{\partial \Psi}{\partial t} = \left [\frac{1}{2m} \left ( \frac{\hbar}{i}\vec{\nabla}-\frac{q}{c} \vec{A} (\vec{r}, t) \right ) ^2 +q\phi (\vec{r}, t)\right ]\Psi$
$-\frac{\hbar}{i}\frac{\partial \Psi}{\partial t} = \left [ \frac{1}{2m}\left ( -\hbar ^2 \nabla ^2-\frac{\hbar q}{ic}\left ( (\nabla A) + (A \nabla)\right ) +\frac{q^2}{c^2}A^2\right ) +q\phi\right ]\Psi$

15. Jun 27, 2014

### ChrisVer

you must understand that it's not a matter of "definition", but a simple fact that you are working with operators. You have an operator $\hat{O}$
which in general doesn't commute with what is right to it. In this case you have partial derivatives of a function which depends on position $\vec{A}$.
And you have that operator acting on the wavefunction after that...so in lines:
$\hat{O}_{x} A(x) \psi(x)$
If the operator $\hat{O}_{x} \equiv \frac{\partial}{\partial x}$
what will you have?

$\frac{\partial}{\partial x} A(x) \psi(x)$

$\frac{\partial A(x)}{\partial x} \psi(x) + A(x) \frac{\partial \psi(x) }{\partial x} = (\frac{\partial A(x)}{\partial x}+ A(x) \frac{\partial}{\partial x}) \psi(x)$

If you write $\hat{G} \psi$ , then the operator $\hat{G}=\frac{\partial A(x)}{\partial x}+ A(x) \frac{\partial}{\partial x}$
And this relation holds in general, even if $O,A$ commutes...in that case the first term in G would be zero. That term corresponds to the commutator of $[O,A]$ of O as I wrote above. $O A \psi = A O \psi + [O,A] \psi = (A O + [O,A]) \psi$
In the case you set $A= ∇$ you have $∇∇$ and the two operators commute- and no extra term has to appear.

(eg another more complicated would be to try to find the $[ ∇^{2}, A(x) ]$

$∇^{2} Ag = ∇∇(Ag) = ∇ [ (∇A) g + A (∇g) ]$
$= (∇^{2}A) g + (∇A) (∇g) + (∇A)(∇g) + A (∇^{2}g)$

so as an operator, taking g out:

$∇^{2}A = (∇^{2}A) + 2 (∇A) ∇ + A ∇^{2}$
or that

$[ ∇^{2}, A(x) ] = (∇^{2}A) + 2 (∇A) ∇$

Last edited: Jun 27, 2014
16. Jun 27, 2014

### ChrisVer

I know that this might seem confusing, or tedious work, but once you do some scholastic calculations with it,, being aware and careful at each step, it's easy to "grab the sense" and it'll be faster for you next time. But you have to do them, because it's a matter of understanding operators and how they act or how you can treat them...as you may already know the result of multiplying two operators is not as it may seem at first... and the confusion is because of the commutation relations.
I guess that since you are dealing with the Electromagnetism Hamiltonian you have already seen the Hamiltonian of the Harmonic Oscillator in the form of the ladder operators and how you find not $H= \alpha^{\dagger} \alpha$ but you also have an additional $+\frac{1}{2}$ term... Although $\alpha= a x + i c p$ and $\alpha^{\dagger}= ax - i c p$ with $a,c$ chosen appropriately...
The same calculations apply in showing the Hamiltonian form from them...

$\alpha^{\dagger} \alpha = (ax - i c p )(a x + i c p) \ne a^{2} x^{2} + c^{2} p^{2}=H$
But
$\alpha^{\dagger} \alpha = a^{2} x^{2} + c^{2} p^{2} + iac xp - iac px = H + iac [x,p]$

NEVERMIND, for your question:

$(a ∇+b A) ( a ∇ + b A) \psi$

To make more obvious what the 1st parenthesis operator acts on:

$(a ∇+b A) ( a ∇\psi + b A \psi)$

multiplying each one with the other:

$a^{2} ∇^{2} \psi + b^2 A^2 \psi + ab ∇(A \psi) + ba A (∇\psi)$

Using Leibniz Rule for the derivative in the 3rd term:

$a^{2} ∇^{2} \psi + b^2 A^2 \psi + ab (∇A) \psi + ab A (∇ \psi) + ba A (∇ \psi)$

adding the same terms:

$a^{2} ∇^{2} \psi + b^2 A^2 \psi + ab (∇A) \psi + 2 ab A (∇ \psi)$

taking out $\psi$ since it's all in the right:

$(a^{2} ∇^{2} + b^2 A^2 + ab (∇A) + 2 ab A ∇ ) \psi$

Identifying the operator I had to act on $\psi$ at first with the one I extracted in the last step:

$(a ∇+b A) ( a ∇ + b A)= a^{2} ∇^{2} + b^2 A^2 + ab (∇A) + 2 ab A ∇$

In your case $a= \frac{h}{i}$ and $b= -\frac{q}{c}$

$(\frac{h}{i} ∇- \frac{q}{c} A)^{2} = (\frac{h}{i})^{2} ∇^{2} + (\frac{q}{c})^2 A^2 - \frac{hq}{ic} (∇A) - \frac{2hq}{ci} A ∇$

$(\frac{h}{i} ∇- \frac{q}{c} A)^{2} = -h^{2} ∇^{2} + \frac{q^{2}}{c^{2}} A^2 - \frac{hq}{ic} [ (∇A) + 2 A ∇ ]$

Last edited: Jun 27, 2014
17. Jun 28, 2014

### carllacan

I think my problem is that I lack a proper way of denoting what is an operator and what is a vector or a derivative. Let me ask a general question:

If we were to denote "chained" operators like $\hat{A_1}(\hat{A_2}(...(\hat{A_n}\Psi)...) ...))$ when is it appropriate to take out the hats anI would say the correct way of solving that would be to apply the transformations of the first operator, then the second, then the third... until we have just the wavefunction or derivatives of it along with some other terms. Is this right?
d apply the transformations corresponding to the operators (derivative, multiplication...)?

I would say the correct way of solving that would be to apply the transformations of the first operator, then the second, then the third... until we have just the wavefunction or derivatives of it along with some other terms. Is this right?

Also, for the case we are on: if we write ∇∇ in component form $\partial q_i(\partial q_j \Psi)$ (with summation over i and j) we will have $\partial q_i(\partial q_j \Psi) = (\partial q_i\partial q_j)\Psi + \partial q_j \partial q_i(\Psi) = 2\partial q_j \partial q_i(\Psi) = 2∇²$. I guess I did something wrong somewhere, but I don't know where.

I can actually move forward without understanding this, but I'd like to get my head around it.

PD: Thank you very much for taking the time to write those answers.

18. Jun 28, 2014

### ChrisVer

The operators with the wavefunctions as that act on what is right to them.
When you have something of the form :
$A_{1} A_{2} f(x)$
Then it means that $A_{1}$ acts on the result you have from $A_{2}f(x)$

$A_{1} (A_{2} f(x))$

To generalize it if you have:
$A_{n} A_{n-1} .... A_{2} A_{1} \psi(x) = A_{n} (A_{n-1} (.... (A_{2} (A_{1} \psi(x) )))...))$
So $A_{2}$ will have to act on your result of $A_{1} \psi(x)$, then the result is acted on by $A_{3}$ then the result is acted on blah blah blah... until you'll get the result of $A_{n-1} G(x)$ (G(x) the general result you achieved from the previous actions) on which you'll act with $A_{n}$.

In case $A_{1}=A_{2}$ (or a more general statement is that if $A_{1},A_{2}$ commute), then:

$A_{1} A_{2} f(x) = A_{2} A_{1} f(x)$

In your case the main thing is that you don't have a commutor:

$\partial_{i} \partial_{j} \psi$
you can interchange the two partial derivatives without any extra term appearing because $[ \partial_{i}, \partial_{j}]=0$.
$\partial_{j} \partial_{i} \psi$

As I said, the extra terms is not just a definition so that you can apply them any where- they are a result of the operators not commuting. In the EM case you have $\partial$ acting on a function of $x,~~A(x)$... So they don't really have to commute (as the mommentum operator p -partial derivative wrt to x, in position representation- does not commute with position x- function of x, A(x)=x , in position repr).

Now going back to my $A_{1,2}$ operators. If they didn't commute:
$A_{1} A_{2} \psi = A_{2} A_{1} \psi + [A_{1},A_{2}] \psi$
you can see that immediately by replacing the commutator.
$A_{2} A_{1} \psi + [A_{1},A_{2}] \psi = A_{2} A_{1} \psi + A_{1}A_{2} \psi - A_{2}A_{1} \psi = A_{1}A_{2} \psi$

This is general, when $A_{1},A_{2}$ commute, then the extra term $[A_{1},A_{2}]=0$ and so it doesn't appear... that's why it doesn't with the partial derivatives you gave before...Because:
$\frac{\partial^{2} f(x)}{\partial x \partial x} =\frac{\partial^{2} f(x)}{\partial x \partial x}$
or to make ti more obvious:
$\frac{\partial^{2} f(x,y)}{\partial x \partial y} =\frac{\partial^{2} f(x,y)}{\partial y \partial x}$

Last edited: Jun 28, 2014
19. Jun 28, 2014

### ChrisVer

Suppose you have a vector:
$x= \begin{pmatrix} x_{1}\\ x_{2}\\ \end{pmatrix}$

On which you act on with the two matrices.....
$A_{1}= \begin{pmatrix} 1 & 0\\ 0 & -1 \\ \end{pmatrix}$

And
$A_{2}= \begin{pmatrix} 0 & 1\\ 1 & 0 \\ \end{pmatrix}$

What's the result of:
$A_{1} A_{2}$

What's the result of:
$A_{2} A_{1}$

Do $A_{1,2}$ commute?
The fact that they don't commute means that their action on $x$ depends on the order they act on... $A_{1}A_{2} x \ne A_{2} A_{1} x$
The extra term that appears is their commutator...
$A_{1}A_{2} x = A_{2} A_{1} x + [A_{1},A_{2}] x$
you can put in the matrices and reconfirm the above result.

Suppose now, instead of $A_{1}$ as given above, you write 1 in the place of -1, then what's
$A_{1} A_{2}$
and
$A_{2} A_{1}$
?
Do they commute?
You can then use:
$A_{1} A_{2} x = A_{2} A_{1} x + [A_{1},A_{2}]x$
and show
$A_{1} A_{2}x= A_{2} A_{1} x$

20. Jun 28, 2014

### Fredrik

Staff Emeritus
21. Jun 28, 2014

### carllacan

I know all that, I just can't make it fit with the ∇∇ case when I write it explicitly $\partial q_i(\partial q_j \Psi) = (\partial q_i\partial q_j)\Psi + \partial q_j \partial q_i(\Psi) = 2\partial q_j \partial q_i(\Psi) = 2∇²$.

But since it must be that I'm just missing some stupid detail I'm gonna leave it for now and I will tackle this another time.

Thank you very much, ChrisVer, for being so patient with me. I really appreciate it.

Going back to the problem. If I state the Schrödinger equation with the expanded Hamiltonian I get:

$i\hbar\frac{\partial \Psi}{\partial t} = \frac{1}{2m}( -h^{2} ∇^{2} + \frac{q^{2}}{c^{2}} A^2 - \frac{hq}{ic} [ (∇A) + 2 A ∇ ] )\Psi + q \phi \Psi$
$\frac{\partial \Psi}{\partial t} = \frac{1}{2i\hbar m}( -h^{2} ∇^{2} + \frac{q^{2}}{c^{2}} A^2 - \frac{hq}{ic} [ (∇A) + 2 A ∇ ] )\Psi + \frac{q \phi}{i\hbar} \Psi$

And for the conjugate
$\frac{\partial \Psi ^*}{\partial t} = \frac{-1}{2i\hbar m}( -h^{2} ∇^{2} + \frac{q^{2}}{c^{2}} A^2 + \frac{hq}{ic} [ (∇A) + 2 A ∇ ] )\Psi ^*- \frac{q \phi}{i\hbar} \Psi^*$
$\frac{\partial \Psi ^*}{\partial t} = \frac{1}{2i\hbar m}( h^{2} ∇^{2} - \frac{q^{2}}{c^{2}} A^2 - \frac{hq}{ic} [ (∇A) + 2 A ∇ ] )\Psi ^*- \frac{q \phi}{i\hbar} \Psi^*$

So now the charge density derivative looks like
$\frac{\partial \rho}{\partial t} = q \frac{\partial}{\partial t}\vert \Psi ^2 \vert =q \frac{\partial}{\partial t}\Psi ^* \Psi = q\left ( \frac{\partial\Psi ^*}{\partial t} \Psi + \frac{\partial\Psi}{\partial t}\Psi^* \right)$

Now I have some doubts. If I put in the above expressions I will get terms like $\Psi\frac{-q^2}{c^2}A^2\Psi^*\Psi$ and$\Psi\frac{q^2}{c^2}A^2\Psi\Psi^*$ which cancel, but also terms like $-\hbar^2 \nabla ^2\Psi ^* \Psi$ and $\hbar^2 \nabla ^2\Psi \Psi^*$, which I'm not sure how to work with. Can I conmute the wavefunctions on the second one so that both terms look the same and I can cancel them, or does the ∇² operator apply before and therefore I have $-\Psi \nabla ^2\Psi^*$ and $\Psi^* \nabla ^2\Psi$?

22. Jun 29, 2014

### ChrisVer

you get terms like:
$\psi^* [\operators] \psi$
In general the current is something which has:
$\psi^* \hat{p}' \psi - \psi \hat{p}' \psi^*$
to this form you are supposed to reach. This form has also some physical meaning. This thing happens to give you $\partial_{t} \rho = ∇ J$
where $\hat{p}'$ is the canonical momentum... in case of your equation, it's not only the derivative, but also has to be modified in order to contain the $A$. By checking the Schroedinger's equation as you gave, you can immediately see that $\hat{p}'= \frac{h}{i} \partial + \frac{q}{c} A$

again you must see what the operator you are using on acts on... $\partial_{t} \psi$ is a function, not operator.

Have you ever worked in finding the current of the normal shroedinger equation-with no potential.

Last edited: Jun 29, 2014
23. Jun 29, 2014

### ChrisVer

Why do you write the 1st step in the equation?

the $\partial q_j \psi$ is a function which you then take the derivative of $\partial q_i$
So what's the meaning of the 1st step?
Instead of writing it as such why don't you write it like:

$\frac{\partial}{\partial q_i} \frac{\partial}{\partial q_j} \psi$

which is by calculus:

$\frac{\partial^{2}}{\partial q_i \partial q_j} \psi$

Last edited: Jun 29, 2014
24. Jun 29, 2014

### carllacan

Because it acts on both the derivative operator and the wavefunction, so I use the pdoruct rule.
No, I actually haven't. I'll look up some examples of that.

Looking at that form for the current: should I have put the wavefunctions like this?

$\frac{\partial \rho}{\partial t} = q\left ( \frac{\partial\Psi ^*}{\partial t} \Psi + \Psi^*\frac{\partial\Psi}{\partial t} \right)$

$\frac{\partial \Psi ^*}{\partial t} = \Psi ^*\frac{1}{2i\hbar m}( h^{2} ∇^{2} - \frac{q^{2}}{c^{2}} A^2 - \frac{hq}{ic} [ (∇A) + 2 A ∇ ] )- \Psi^*\frac{q \phi}{i\hbar}$

(Rather than how I wrote them here, when I took the conjugate of the wavefunction derivative and derived the product in the charge density)
From this I think I can reach an expression like yours.

25. Jun 29, 2014

### ChrisVer

have you ever used in calculus that:

$\frac{\partial}{\partial x} \frac{\partial}{\partial y} F(x,y) = (\frac{\partial}{\partial x}\frac{\partial}{\partial y}) F(x,y) + \frac{\partial}{\partial y} [\frac{\partial}{\partial x} F(x,y)]$

http://en.wikipedia.org/wiki/Product_rule

If you haven't tried that for the easiest thing, I'd hardly recommend that you try it before going in more general stuff...
http://www.physics.ucdavis.edu/Classes/Physics115A/probcur.pdf

From eq (3) to (4) you have to take out one ∇, by the same product rule...
$f^* (x) ∇^{2} f(x) = ∇ ( f^* (x) ∇f(x) ) - (∇f^* (x) ) (∇f(x))$
the 2nd term in the above, cancels out by a similar appearing from the other term, and the 1st one together with one more coming from the other term again will give you something like:
$∇ (f^* (x) ∇f(x) - f(x)∇f^* (x) )$
And you get the probability current once you identify the above with:
$∇J$

Last edited: Jun 29, 2014
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