Quantum particle in a magnetic field

[carllacan]

Homework Statement

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 $\vec{j}(\vec{r},t)$

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

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?

 

[carllacan]

Question edited with more information.

 

[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...

 

[carllacan]

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...

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?

 

[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.

 

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

 

[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?

 

[carllacan]

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?

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?

 

[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
$J= Something$

 

[ChrisVer]

editted^

 

[carllacan]

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

 

[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?

 

[carllacan]

$(c ∇+ d A)(c ∇+dA)= c^{2} ∇^{2} + d^{2}A^{2} + cd A∇ + cd ∇A$
in the 3rd term, nabla acts on the wavefunction alone, in the 4th term nabla acts on both A and the wavefunction:
$(c ∇+ d A)^{2}= c^{2} ∇^{2} + d^{2}A^{2} + cd A∇+ cd A∇+ cd (∇A)$

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 $c^{2} ∇^{2} =c^{2} ∇∇+ c^{2} (∇∇)$?

 

[carllacan]

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

 

[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

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

 

[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 ∇ ]$

 

[carllacan]

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

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

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

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.

 

[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:
$\frac{\partial^{2} f(x,y)}{\partial x \partial y} =\frac{\partial^{2} f(x,y)}{\partial y \partial x}$

 

[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
$A_{1} A_{2}x= A_{2} A_{1} x$

 

[Fredrik]

I think this thread should have a link to carllacan's other thread about a similar topic. It's about some of the details of the calculations here. https://www.physicsforums.com/showthread.php?p=4785337#post4785337.

 

[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 going to 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 $\Psi^* \nabla ^2\Psi$?

 

[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.

 

[ChrisVer]

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

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:

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

 

[carllacan]

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:

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

Because it acts on both the derivative operator and the wavefunction, so I use the pdoruct rule.

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

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.

 

[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:
$∇J$

 

[Fredrik]

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

it acts on both the derivative operator and the wavefunction, so I use the pdoruct rule.

The product rule says that $(fg)'=f'g+fg'$, or equivalently (just a different notation), that $D(fg)=(Df)g+f(Dg)$. It doesn't say that $D^2f=(D(D))f+D(Df)$. This right-hand side doesn't make sense. The domain of the derivative operator $D$ is the set of differentiable functions. $D$ is not an element of that set. Also, $D^2$ is by definition the product $DD$, and products of linear operators are defined by $(AB)f=A(Bf)$. So $D^2f=D(Df)$. If you write a second term on the right-hand side, we no longer have an equality (unless that extra term is zero of course).

The reason why you have to do something that looks like an application of the product rule in the calculation $(DA(x))f=D(A(x)f)=A'(x)f+A(x)Df$ is that $D$ doesn't commute with $A(x)$. $A'(x)$ is not the result of $D$ acting on $A(x)$. $A(x)$ is an operator that isn't in the domain of $D$. $A'(x)$ is not the derivative of $A(x)$. $A(x)$ isn't something that can be differentiated. It's just an operator. The operator-valued function $A$ is what can be differentiated. If we also use the symbol D for the derivative operator on the set of differentiable operator-valued functions. Then we can write $A'=DA$, and therefore $A'(x)=(DA)(x)$, but then we'd have $(DA(x))f=D(A(x)f)=(DA)(x)f+A(x)Df$, and this isn't quite the product rule. The "product" $A(x)f$ isn't even a product. It's the value of $A(x)$ at $f$. And even if we view it as a product and naively "use the product rule", we'd have the nonsensical $D(A(x)f)=D(A(x))f+A(x)Df$ instead of the correct $D(A(x)f)=(DA)(x)f+A(x)Df$.

 

[ChrisVer]

$A(x)$ as something acting on the position-representation and depends only on $x$ can be viewed as a function of $x$ and so is differentiable by the partial derivatives. But not to be misunderstood - of course as an object alone, it's an operator.

That's generally why, in the photon field quantization you can have quantities like:
$[A,\dot{A}] \ne 0$
while for two functions in general you'd expect them to commute ($f(x)g(x)=g(x)f(x)$).
Or see the above as the canonical commutation relation.

Also in general, for the current problem, you don't have to care about $\phi$ potential field... It's either going to give you a fixed value for your energy, or you can gauge it out by choosing a gauge (eg the Lorentz gauge which says that $\partial_{\mu} A^{\mu}=0$ where:
$A^{\mu}=( \phi, \vec{A})^{T} )$
And so you can relate $\phi$ with $\vec{A}$ and absorb it in a new definition of $A$, erasing it from your Schr. equation.

 

[Fredrik]

$A(x)$ as something acting on the position-representation and depends only on $x$ can be viewed as a function of $x$

Right, but to say that A(x) is a function of x, is to say that A is a function, and that A(x) is an element of the range of that function. To be more specific, A is an operator-valued function and for each x, A(x) is an operator.

Note that the "product" that we're dealing with is something like $A(x)f$, not $Af$ or $A(x)f(x)$. This makes it a bit of a stretch to say that we need to use the product rule because A(x) is a function of x.

I'm actually very puzzled about what $A(x)$ really means here. Is $A$ a function defined by the classical version of Maxwell's equations, and then we somehow convert that to a function that takes operators as input and feed it a triple of position operators as input? (In the other thread I showed in a very non-rigorous way that this appears to imply that $D(A(x)f)=A'(x)f+A(x)Df$). Or is it an operator-valued function defined in an entirely different way? Maybe it's the former in quantum mechanics, and the latter in QED? Hm, quantum fields are supposed to be operator-valued distributions, so it's probably even worse than that actually.

 

[ChrisVer]

In Schrodinger's equation, $A$ is the normal (classical) magnetic field. And that can be viewed from the fact that indeed you can apply the gauge fixing $\partial A=0$- exactly because it's classical.
In case you work in the QFT framework, then things differ, exactly because you start seeing $A$ as a field (so you start imposing commutation relations on it), and thus the gauge fixing stops working ($\partial A \ne 0$ as an operator relation). In that case you insert the above as an operator relation, and impose that the mean value of it is indeed zero:
$< \phi | \partial A | \phi> =0$
as the gauge fixing.
Maybe from this you can see the "operator-valued distribution" behind since we speak for mean-value)
But you can always see things like operators and commutations, it doesn't destroy anything to do so...
For example a correct way -as I said above- to find $[x,p]$ although $x$ is an operator, in position space it becomes just a function of x $f(x)=x$.
And you can apply:
$-ih x \partial_{x} \phi +ih \partial_{x} (x \phi) = ih \phi$
$[x,p]=ih$

I am not really sure but you can say that $A(x)= <x|A|x>$ and happens to be a normal function of $x$ as is $psi$- which in general can be given by $<x|\psi>$ and $|\psi>=creations|0>$.

 

[Fredrik]

If we define A as a real-valued function on $\mathbb R^3$, through the formula A(x)=<x|A|x> or otherwise, and B is an operator, then A(x)+B would be interpreted as A(x)I+B, where I is the identity operator. If A(x) is a number times the identity operator, then A(x) would commute with every operator. In particular, we would have (DA(x))f=A(x)Df.

That's why I'm suggesting that maybe we can "convert" A to a function that takes triples of operators to operators, and then input the three position component operators. The conversion could perhaps be a rigorous version of this: $A(\hat x)=\int d^3x A(x)|x\rangle\langle x|$.

 

[ChrisVer]

the operator in position space is just the partial derivative, and partial derivative will act on $\mathbb{R}^{3}$ functions such as $A$, as it does on wavefunctions too, because you have x-dependence.
You can't really say that :
$\partial G(x) \psi(x) = G(x) \partial \psi(x)$ can you?
Check the [x,p] I gave above.

In your case then, still the identity remains untouched- as it should be- but the elements are acted on.

Of course you can work in any representation you like, and get the result, but the result is obvious in position repr, where you have to take just the derivative of a function.

With the identity thing then, I find it quite difficult to see the $[x,p]$
$[x,p]= xp- px$
if I write $f(x)=x I$ and follow your logic - identity will commute with p- I will get another commutation relation:
$[x,p]= xI p - xI p =0$

 

[Fredrik]

the operator in position space is just the partial derivative, and partial derivative will act on $\mathbb{R}^{3}$ functions such as $A$, as it does on wavefunctions too, because you have x-dependence.

I disagree. Consider $(\nabla+A(x))^2$. We have
\begin{align}
(\nabla+A(x))^2=(\nabla+A(x))_i(\nabla+A(x))_i =(\partial_i+A_i(x))(\partial_i+A_i(x)) =\partial_i\partial_i+\partial_i A(x)+A(x)\partial_i+A_i(x)A_i(x)
\end{align} Now let's just consider the second term. It's a product of two operators, and products are defined by $(AB)f=A(Bf)$ for all f. So for all f, we have
$$(\partial_i A_i(x))f=\partial_i(A_i(x) f).$$ What this means is that for all y, we have
$$\big((\partial_i A_i(x))f\big)(y) = \partial_i(A_i(x) f)(y).$$ The right-hand side is the value at y of the partial derivative of the function A_i(x)f, i.e. the function $z\mapsto (A_i(x)f)(z)$. It's not equal to
$$\frac{\partial}{\partial x_i}\big(A_i(x)f(x)\big)$$ so the product rule doesn't apply. Sure, we have "x dependence", but we would need "y dependence".

To evaluate $(\partial_i A_i(x))f$, we have to know the proper way to make sense of $A_i(x)$. This is above my pay grade. In the other thread, I made a very non-rigorous argument for why we should have $(\partial_i A_i(x))f =(\partial_i A_i)(x)f +A_i(x)f$ when $A_i$ is an operator-valued function. Obviously, the only thing non-rigorous arguments are good for is that they can help us guess what definitions may be useful and what statements we should try to prove.

With the identity thing then, I find it quite difficult to see the $[x,p]$
$[x,p]= xp- px$
if I write $f(x)=x I$ and follow your logic - identity will commute with p- I will get another commutation relation:
$[x,p]= xI p - xI p =0$

That x is definitely the position operator, defined by $(\hat x f)(x)=xf(x)$ for all f and all x. It's not a real-valued function times the identity operator.

 

[Oxvillian]

I haven't been through the whole thread in detail, but it seems to me that maybe there's some confusion about operators and their representatives in the x-representation.

Consider the following x-representation calculation where $X$=position operator and $x$=plain old x-coordinate:
$$\begin{eqnarray} \langle x \lvert PA(X) \vert \psi \rangle &=& \int \langle x \lvert PA(X) \lvert x' \rangle \langle x' \lvert \psi \rangle dx'\\ &=& \int \langle x \lvert PA(x') \lvert x' \rangle \langle x' \lvert \psi \rangle dx'\\ &=& \int A(x') \langle x \lvert P \lvert x' \rangle \langle x' \lvert \psi \rangle dx'\\ &=& \int A(x') [-i\hbar \delta'(x-x')] \langle x' \lvert \psi \rangle dx'\\ &=& -i\hbar \frac{d}{dx}[A(x)\psi(x)] \end{eqnarray}$$
The moral of the story being that although the final rule involves differentiating the function $A(x)$, at no point do we ever have to worry about "differentiating operators".

 

[ChrisVer]

I don't really understand what you are struggling to say Fredrik.. sorry...
Why would you write for a function:
$A(x) f(x) = h(y)$
And not just say that:
$A(x) f(x)= h(x)$
?

The partial derivative then $\partial h(x) = h'(x)$ remains a function of x...

Maybe we are losing it in $A(x)$... but in my viewpoint, when $A(\hat{x})$ will act on $\psi(x)$, you will have just $A(x)$ a function.
If you have problems with $A(x) I$ and it being able to commute with all operators... well this is somewhat wrong. If your operator has the property to change the spatial component, then that's know right, obviously...why? because your final function, which $\partial$ will act on is $h(x)$
the derivative is just the:

$\lim_{\epsilon \rightarrow 0} \frac{ h(x+ \epsilon) - h(x)}{\epsilon}$

$=\lim_{\epsilon \rightarrow 0} \frac{ A(x+ \epsilon) f(x+ \epsilon) - A(x) f(x)}{\epsilon}$

$=\lim_{\epsilon \rightarrow 0} \frac{ [A(x)+ \epsilon A'(x)] [f(x)+ \epsilon f'(x)] - A(x) f(x)}{\epsilon}$

$=A'(x) f(x) + A(x) f'(x)$

If by any means you had $A(z)$ with z independent of x, then the partial derivative wouldn't "see" that, and it would only act on $f(x)$

 

[Fredrik]

I don't really understand what you are struggling to say Fredrik.. sorry...
Why would you write for a function:
$A(x) f(x) = h(y)$
And not just say that:
$A(x) f(x)= h(x)$
?

(Just a reminder, both to myself and to you. What we're discussing here is the case where the $A_i$ are operator-valued functions with domain $\mathbb R^3$, and whether it makes sense to say that we need to use the product rule because $A_i(x)$ depends on $x$. My position is that it doesn't).

I'm not doing anything like A(x)f(x)=h(y). What I'm doing is to say that for all functions f and g, if f=g, then f and g have the same domain, and for all y in that domain, we have f(y)=g(y).

x is an element of some set S, such that for each $x\in S$, we're dealing with two functions, I'll call them $f_x$ and $g_x$ here, to keep the notation simple. The x in the equality $f_x=g_x$ is a variable that represents some element of the set S. Regardless of what element that is, the equality means that $f_x$ and $g_x$ have the same domain, and for all $y$ in that domain, we have $f_x(y)=g_x(y)$. $y$ is of course a dummy variable here. It can be replaced by any other symbol, except $x$, which is already reserved to denote an element of S.

The x in the equality $(\partial_i A_i(x))f=\partial_i(A_i(x) f)$ only determines which functions we're dealing with. On the right-hand side, x is not the input to the function on which $\partial_i$ acts. It's a variable whose value determines which function the $\partial_i$ acts on. On the left-hand side, the value of x determines which operator $\partial_i$ is multiplied with.

Maybe we are losing it in $A(x)$... but in my viewpoint, when $A(\hat{x})$ will act on $\psi(x)$

$\psi(x)$ is just a number. I assume that you mean $\psi$. $\psi(x)$ is an element of the range of $\psi$. I think that if you want to sort this out, you will have to be very careful with the terminology, so that you never end up thinking of a number as a function. Also, be careful with "for all" statements.

I am extremely nitpicky with these things. I never refer to $x^2$ or $f(x)$ as a "function" for example. Some people think that's ridiculous, but it's really paying off in problems like this.

Regarding $A(\hat x)$. I know two non-rigorous ways to deal with that, and they both yield the desired result. One way is covered in the other thread. The other is by the methods of Sakurai's book, as in Oxivillian's post. It's $A(x)$ (with x a real number or a triple of real numbers) that I think will give us a calculation that doesn't even look like the product rule.

Edit: Something to think about: How would you define the product of $\partial_i$ and $A_i(x)$? Is it not through the usual $(AB)f=A(Bf)$? That definition tells us that $(\partial_i A(x))f=\partial_i(A_i(x)f)$. To think that the $\partial_i$ has anything to do with the symbol x here is to confuse the function $A_i(x)f$, which takes an arbitrary y to $(A_i(x)f)(y)$, with the function-valued function $x\mapsto A_i(x)f$, or to first confuse f with f(x), and then interpret $\partial_i(A_i(x)f(x))$ (which is not what we're dealing with) as the ith partial derivative at x of the function $y\mapsto A_i(y)f(y)$. (I'm using y here, because it's potentially confusing to use the already reserved symbol x as a dummy variable).