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- TL;DR Summary
- Is the dot product of two vector operators only understood to be the sum of the squares of the components when the vector operators are written with their components in Cartesian co-ordinates? (See main text for more).

Hi. I hope everyone is well. I'm just an old person struggling to make sense of something I've read and I would be very grateful for some assistance. This is one of my first posts and I'm not sure all the LaTeX encoding is working, sorry. Your help pages suggested I add as much detail as possible but I'm not sure how much is too much, so I'll apologise for that as well. If you get bored, skip it, or perhaps write a reply suggesting that I break the question down into multiple smaller posts etc. In Quantum Mechanics (with 3-spatial dimensions) the angular momentum is often usefully expressed in spherical co-ordinates rather than Cartesian co-ordinates, so I'll take that as an example:

This is the (vector) Angular Momentum Operator (for a single particle with some arbitrary wave function)

## \hat{L} ## = ## \underline{e_x} \hat{L_x} + \underline{e_y} \hat{L_y} + \underline{e_z} \hat{L_z} ##

The components ##\hat{L_x} ## etc. are themselves operators but they are just scalar operators (and represent the angular momentum in the x, y, z directions as usual). We have expressions for those scalar operator components in terms of multiplication by the cartesian co-ordinates and derivatives with respect to the cartesian co-ordinates. I haven't written them out - but I'm sure you know what they are. Please consider [Equation 1] to be using Cartesian co-ordinates throughout.

The (vector) angular momentum operator can also be written as follows:

## \hat{L} ## = ## -i \hbar \underline{e_{\phi}} \frac{\partial}{\partial \theta} \, + \, i \hbar \underline{e_{\theta}} \frac{1}{sin \theta} \frac{\partial}{\partial \phi} ##

Now, I'm sure you're also familiar with the (scalar) operator ## \hat{L}^2 ##. Using [Equation 1] and the usual procedure for the dot product of two vector operators ## \hat{L} . \hat{L} ## we have the following:

## \hat{L}^2 \, = \, \hat{L_x}^2 + \hat{L_y}^2 + \hat{L_z}^2 ##

We have expressions for the (scalar) operators ## \hat{L_x} ## etc. which are the components of that vector operator. So we can put them in and get a final expression for a scalar operator ##L^2## in terms of (partial) derivatives with respect the cartesian co-ordinates x,y, z. That's done in many places (references can be provided if required - but basically any textbook or reliable online resource). Then using the chain rule for functions of many variables we can re-write that expression in terms of (partial) derivatives with respect to spherical co-ordinates r, ##\theta## , ##\phi## . This is what you get:

$$ \hat{L}^2=-\hbar^2\left[\frac{1}{\sin \theta} \frac{\partial}{\partial \theta}(\sin \theta \frac{\partial}{\partial \theta}) + \frac{1}{\sin^2 \theta} \frac{\partial^2}{\partial \phi^2} \right] $$

However, if you use [Equation 2] and apply the usual procedure for the dot product of two vector operators you obtain (well...I would have thought...) the following:

## \hat{L}^2 = \hbar^2 (\frac{\partial^2}{\partial \theta^2} + \frac{1}{sin^2 \theta} \frac{\partial^2}{\partial \phi^2}) ##

Being positive about things, that is almost the same as [Equation 3]. Being sensible, it's just not the same which is all that matters.

I spent about a day being stuck on this. WHY IS IT NOT THE SAME? I'd appreciate some advice just on that issue.

OK... that's probably already far too much. I was going to talk about what I think is the problem... very briefly.... Looking carefully at the dot product of the vector operators ## \hat{L} . \hat{L} ## when written in cartesian co-ordinates we only have terms like (unit vector) d/dx (unit vectors times a function) which is fine because those unit vectors are fixed and unchanging, so pass through that derivative like constants. However, when the vector operators are written out in spherical co-ordinates then the dot product can have terms like (unit vector) ##d/d{\theta}## (unit vectors times a function) but those unit vectors are not unchanging with respect to the spherical co-ordinates like ##\theta##, so the terms to the right of the derivative operator must be differentiated as a product. To say it another way, the component operators can sometimes act on the unit vectors and then the dot product is complicated. More-over, can we really only compute the dot product of two vector operators by first making sure they are written as vector operators with cartesian components?

Thank you for your time, what can I say? You've done well just to read this far.

This is the (vector) Angular Momentum Operator (for a single particle with some arbitrary wave function)

## \hat{L} ## = ## \underline{e_x} \hat{L_x} + \underline{e_y} \hat{L_y} + \underline{e_z} \hat{L_z} ##

[Equation 1]

It has been written as usual with Cartesian co-ordinate components ( just to be clear then, ## e_x ## , ## e_y ##, ## e_z ## are unit vectors in the x, y, z direction and I haven't put hats over them because I'm saving the hats for operators like ##L_x ##. I have underlined those vectors but the underline doesn't seem to be displaying in my preview, sorry).The components ##\hat{L_x} ## etc. are themselves operators but they are just scalar operators (and represent the angular momentum in the x, y, z directions as usual). We have expressions for those scalar operator components in terms of multiplication by the cartesian co-ordinates and derivatives with respect to the cartesian co-ordinates. I haven't written them out - but I'm sure you know what they are. Please consider [Equation 1] to be using Cartesian co-ordinates throughout.

The (vector) angular momentum operator can also be written as follows:

## \hat{L} ## = ## -i \hbar \underline{e_{\phi}} \frac{\partial}{\partial \theta} \, + \, i \hbar \underline{e_{\theta}} \frac{1}{sin \theta} \frac{\partial}{\partial \phi} ##

[Equation 2]

This is using spherical co-ordinates throughout. The vector operator is given in terms of components along the unit spherical co-ordinate vectors (just to be clear again: ## e_{\theta} ## , ## e_{\phi} ## are unit vectors in the ## \theta ## = azimuthal angle and ## \phi ## = polar angle directions). Also those components are clearly seen to be (scalar) operators involving only multiplication by the spherical co-ordinates and derivatives with respect to the spherical co-ordinates.Now, I'm sure you're also familiar with the (scalar) operator ## \hat{L}^2 ##. Using [Equation 1] and the usual procedure for the dot product of two vector operators ## \hat{L} . \hat{L} ## we have the following:

## \hat{L}^2 \, = \, \hat{L_x}^2 + \hat{L_y}^2 + \hat{L_z}^2 ##

We have expressions for the (scalar) operators ## \hat{L_x} ## etc. which are the components of that vector operator. So we can put them in and get a final expression for a scalar operator ##L^2## in terms of (partial) derivatives with respect the cartesian co-ordinates x,y, z. That's done in many places (references can be provided if required - but basically any textbook or reliable online resource). Then using the chain rule for functions of many variables we can re-write that expression in terms of (partial) derivatives with respect to spherical co-ordinates r, ##\theta## , ##\phi## . This is what you get:

$$ \hat{L}^2=-\hbar^2\left[\frac{1}{\sin \theta} \frac{\partial}{\partial \theta}(\sin \theta \frac{\partial}{\partial \theta}) + \frac{1}{\sin^2 \theta} \frac{\partial^2}{\partial \phi^2} \right] $$

[Equation 3]

That is quite a lot of work to get the final operator ## L^2 ## in terms of partial derivatives w.r.t. the spherical co-ordinates. It would seem faster to go straight from [Equation 2] where we already had the components as derivatives w.r.t. spherical co-ordinates.However, if you use [Equation 2] and apply the usual procedure for the dot product of two vector operators you obtain (well...I would have thought...) the following:

## \hat{L}^2 = \hbar^2 (\frac{\partial^2}{\partial \theta^2} + \frac{1}{sin^2 \theta} \frac{\partial^2}{\partial \phi^2}) ##

Being positive about things, that is almost the same as [Equation 3]. Being sensible, it's just not the same which is all that matters.

I spent about a day being stuck on this. WHY IS IT NOT THE SAME? I'd appreciate some advice just on that issue.

Anyway... that is the motivation for the original question: If an arbitrary vector operator ## \hat{L} ## is written as follows: ##\hat{L} ## = ## \vec{a} \hat{L_a} + \vec{b} \hat{L_b} ## where (i) ##\hat{L_a}## and ## \hat{L_b}## are scalar operators (representing the components of the vector operator in the a and b vector directions). (ii) You only know that the vectors ##\vec{a} ## and ##\vec{b} ## are orthogonal and unit vectors (but not necessarily unit vectors along Cartesian axis). Then is it impossible to conclude that the operator ## \hat{L} . \hat{L} ## (more commonly written as ##L^2## ) would be equal to the operator that is the sum of the squares of the component operators ## \hat{L_a}^2 + \hat{L_b}^2 ## ? Where, as usual, the square of a scalar operator ##\hat{L_a}^2## and ##\hat{L_b}^2## is understood as the composition of operators, e.g. ##L^2 (\psi) = L (L(\psi)) ## |

Thank you for your time, what can I say? You've done well just to read this far.

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