Notation for the nabla operator arguments

[carllacan]

Hi.

In this development
(c ∇+ d A)(c ∇+dA)= c^{2} ∇^{2} + d^{2}A^{2} + cd A∇ + cd ∇A
(c ∇+ d A)^{2}= c^{2} ∇^{2} + d^{2}A^{2} + cd A∇+ cd A∇+ cd (∇A)

I feel like we have "two" different ∇ operators. At the end of the first line ∇ acts on A and the test function (not shown). At the second line it acts just on the A operator or the test function.

i was wondering what would be the best way to denote this two different "views" on the operator. Should I consider the first one to be a proper operator and the second one as just a 3-vector?

Thank you for your time.

 

[Fredrik]

Is $d$ a number? What exactly is $A$? How did you get different results in the right-hand sides when the two left-hand sides are the same?

The product of two linear operators is defined by $AB=A\circ B$. This means that for all f in the domain of $B$ such that $Bf$ is in the domain of $A$, we have $(AB)f=A(Bf)$.

To know how to handle $(\nabla A)f$, I need to know what $A$ is. If it's the linear operator defined by $(Af)(x)=g(x)f(x)$, where g is some function, then
$$(\nabla A)f=\nabla(Af)=\nabla(gf)=(\nabla g)f+g\nabla f =(\nabla g+g\nabla)f.$$ Here $\nabla g$ is of course just a vector. If you denote the function g by $A$, then this will look confusing. I can only guess that this is what you're doing.

 

[carllacan]

Okay, here is some context: https://www.physicsforums.com/showthread.php?t=754798

A is a vector operator, concretely the magnetic vector potential. The two lines come from this answer: https://www.physicsforums.com/showpost.php?p=4754929&postcount=9

 

[Fredrik]

OK, that specific A is really a triple of operators $(A_1,A_2,A_3)$, so I would suggest that you use a notation that reflects that. For example $\vec A=(A_1,A_2,A_3)$. I guess I should have realized that $A$ is a triple, since $\nabla$ is. The linear combination $c\nabla+dA$ doesn't make much sense otherwise. We can write $\nabla=(\partial_1,\partial_2,\partial_3)$. A linear combination of these triples of operators is a triple of operators. And the "square" of such a triple is a single operator (not a triple) computed using the dot product. For example $\vec A^2=\sum_{i=1}^3 A_iA_i$. So
\begin{align}
&(c\nabla+d\vec A)^2 = \sum_i (c\nabla+d\vec A)_i (c\nabla+d\vec A)_i =\sum_i (c\partial_i+dA_i) (c\partial_i+dA_i)\\
&=\sum_i (c^2\partial_i\partial_i+cd\partial_i A_i+dcA_i\partial_i+d^2A_iA_i)
\end{align} Now if we just use the definitions of addition, scalar multiplication and multiplication of operators, we see that
\begin{align}
(c\nabla+d\vec A)^2 f = \sum_i \big(c^2\partial_i(\partial_i f)+cd\partial_i( A_if)+dcA_i(\partial_i f)+d^2A_i(A_if)\big)
\end{align} The result of that two-line calculation above can also be written as
$$c^2\nabla^2+cd\nabla\cdot\vec A+dc\vec A\cdot\nabla+d^2\vec A^2.$$ If this notation confuses you, don't use it. If you want to use it, the correct way is
$$(c^2\nabla^2+cd\nabla\cdot\vec A+dc\vec A\cdot\nabla+d^2\vec A^2)f =c^2\nabla\cdot(\nabla f)+cd\nabla\cdot(\vec Af)+dc\vec A\cdot(\nabla f)+d^2\vec A\cdot\vec Af).$$ The best way to see that is to just compare this to the less confusing notation above.

 

[carllacan]

\begin{align}
(c\nabla+d\vec A)^2 f = \sum_i \big(c^2\partial_i(\partial_i f)+cd\partial_i( A_if)+dcA_i(\partial_i f)+d^2A_i(A_if)\big)
\end{align} The result of that two-line calculation above can also be written as
$$c^2\nabla^2+cd\nabla\cdot\vec A+dc\vec A\cdot\nabla+d^2\vec A^2.$$

I have a problem here. Wouldn't the $c^2\partial_i(\partial_i f)$ and $cd\partial_i( A_if)$ terms apply differentiation of products and become this?
\partial_i(A_i f) = \partial_i A_i f + A_i \partial_i f
\partial_i(\partial_i f) = \partial_i \partial_i f + \partial_i \partial_i f

 

[Fredrik]

OK, this is much more difficult than I thought at first. In my previous posts, I overlooked the fact that the $A_i$ aren't operators. They're operator-valued functions. But what is the domain of those functions? Can it be $\mathbb R^3$. Then for each $x\in\mathbb R^3$, $A_i(x)$ is an operator. But the product of two operators is still defined by (AB)f=A(Bf), so we would still have $(\partial_i A_j(x))f =\partial_i(A_j(x)f)$, and for all y,
$$((\partial_i A_j(x))f)(y)= \partial_i(A_j(x)f)(y).$$ The $\partial_i$ here should be interpreted as $\partial/\partial y_i$, not $\partial/\partial x_i$, so the product rule doesn't apply. But I don't think your calculation is supposed to look like this. I think it's supposed to include a step that looks like an application of the product rule. This means that the domain of the $A_i$ functions is not $\mathbb R^3$.

Instead, the $A_i$ must be functions that take triples of operators to operators. The x in $A_i(x)$ is the triple $(x_1,x_2,x_3)$ of position operators. That makes it very hard to see how and why we're supposed to "use the product rule".

I'm going to use an extremely non-rigorous argument. I will only consider the 1-dimensional case. Let Q be the position operator, defined by (Qf)(x)=xf(x). Let D be the derivative operator, defined by (Df)(x)=f'(x). You know that if we define P=-iD, we have [Q,P]=i. This implies that [D,Q]=[iP,Q]=i(-i)=1. This implies that $[D,Q^k]=kQ^{k-1}$. Suppose that A can be expressed as a power series, $A(Q)=\sum_k a_k Q^k$. We have
\begin{align}
DA(Q) &=D\sum_k a_k Q^k =\sum_k a_k DQ^k = \sum_k a_k (Q^kD+[D,Q^k]) =\sum_k a_k(Q^kD+(k-1)Q^{k-1})\\
&=A(Q)D+A'(Q).
\end{align} So
$$(DA(Q))f =(A(Q)D+A'(Q))f =A(Q)(Df)+A'(Q)f.$$ It looks like we have applied the product rule, but we really haven't, at least not the standard product rule that we're used to. We have done something rather different, and it's not what I did here either. Some of the things I did are invalid mathematical operations. To write A(Q) as a power series only makes sense when Q is bounded, and our Q isn't. I have also ignored that the domain of $[Q,D]$ can't be the entire Hilbert space. And I don't even want to think about what else in my calculation is just wishful thinking.

I guess you have to either study 800 pages of toplogy and analysis, or just write things down as if you're using the product rule.

 

[Fredrik]

\partial_i(\partial_i f) = \partial_i \partial_i f + \partial_i \partial_i f

This would mean that $\partial_i\partial_i = 2\partial_i\partial_i$, which would mean that $\partial_i\partial_i=0$. Did you mean $(\partial_i(\partial_i))f$ in the first term? I suppose this could appear to make sense at first, but when you think about it, you see that it doesn't. $\partial_i$ acting on $\partial_i$?

Check out the non-rigorous calculation in my previous post. The A'(Q) in $DA(Q) =A(Q)D+A'(Q)$ is the commutator [D,A(Q)]. When you evaluate $\partial_i\partial_j$, you can certainly write it as $\partial_j\partial_i+[\partial_i,\partial_j]$, but the commutator is zero, so there's no additional term.

 

[carllacan]

I guess you have to either study 800 pages of toplogy and analysis, or just write things down as if you're using the product rule.

I guess I'll do the latter for now and try to do the former in the future. Care to recommend some books on those topics? If possible oriented to this kind of applications.

And btw

This means that the domain of the Ai functions is not R3.

It is a "physical" operator, the magnetic vector potential, so I think it is indeed a functions of R3.

 

[carllacan]

This would mean that $\partial_i\partial_i = 2\partial_i\partial_i$, which would mean that $\partial_i\partial_i=0$. Did you mean $(\partial_i(\partial_i))f$ in the first term? I suppose this could appear to make sense at first, but when you think about it, you see that it doesn't. $\partial_i$ acting on $\partial_i$?

An operator acting on an operator. Where's the problem? If it can act on $A_i$ why can't it act on $\partial_i$?

Check out the non-rigorous calculation in my previous post. The A'(Q) in $DA(Q) =A(Q)D+A'(Q)$ is the commutator [D,A(Q)]. When you evaluate $\partial_i\partial_j$, you can certainly write it as $\partial_j\partial_i+[\partial_i,\partial_j]$, but the commutator is zero, so there's no additional term.

I think here would come in handy those 800 pages on topology xD. I'm lost. I'm going to leave it here for now and accept the calculation as it is. Thank you.

 

[Fredrik]

I guess I'll do the latter for now and try to do the former in the future. Care to recommend some books on those topics? If possible oriented to this kind of applications.

Not sure what to recommend. I learned topology from a number of different sources, including Munkres and the appendix of Sunder's book on functional analysis. There may be better options. You can search for topology in the forum section about books. A lot of people recommend Kreyszig for functional analysis, I think mainly because it doesn't require that you know topology already. But I haven't used it myself.

For functional analysis, I have used a lot of books. Sunder is a pretty good choice when you get to Banach and C*-algebras. Not sure what to use for the basics. Probably Kreyszig. Conway is nice in some ways, but it's very difficult to follow his proofs.

You will need a solid foundation in linear algebra. I like Axler for that. The selection of topics is a very close match to what you need for QM.

And btw It is a "physical" operator, the magnetic vector potential, so I think it is indeed a functions of R3.

As I explained, the product rule wouldn't apply then. So I'm pretty sure it takes position operators, not numbers, as input.

An operator acting on an operator. Where's the problem? If it can act on $A_i$ why can't it act on $\partial_i$?

But it's not acting on $A_i$ (which really means $A_i(Q)$). That was one of the points of the non-rigorous calculation. DA(Q) was the product of D and A(Q), defined by $(DA(Q))(f)=D(A(Q)f)$. The right-hand side happens to be equal to $A'(Q)f+A(Q)Df$, so it may look like we have just used the product rule, but we're really doing something else. A(Q) is not in the domain of D. By the way, if D acts on each "factor" as we might expect from the product rule, shouldn't it act on the operator $A(Q)$ rather than on the operator-valued function $A$?

I have to leave the computer for a few hours now.