# Dot product of two vector operators in unusual coordinates

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• Old Person
In summary: And the components ## \hat{L_x} ## etc. are scalar operators involving only multiplication by the cartesian co-ordinates and derivatives with respect to the cartesian co-ordinates, what is the expression for the (scalar) operator ##L^2## in terms of partial derivatives w.r.t. the cartesian co-ordinates?

#### Old Person

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} ##
[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)) ##
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.

Last edited by a moderator:
Spherical coordinates are an example of curvilinear coordinates, where the the coordinate lines do not intersect at right angles. That is why the dot product, which is the projection of one vector on another, depends on the position in coordinate space.

See, e.g., https://en.wikipedia.org/wiki/Curvilinear_coordinates

dextercioby and Old Person
DrClaude said:
Spherical coordinates are an example of curvilinear coordinates, where the the coordinate lines do not intersect at right angles.
??? My impression is that the coordinate lines still intersect at right angles for spherical coordinates. Not that this would be important, but still, no need to make false statements.

Old Person
gentzen said:
??? My impression is that the coordinate lines still intersect at right angles for spherical coordinates.
Except at the coordinate singularities at the two poles, yes.

Old Person and gentzen
DrClaude said:
Spherical coordinates are an example of curvilinear coordinates, where the the coordinate lines do not intersect at right angles. That is why the dot product, which is the projection of one vector on another, depends on the position in coordinate space.

See, e.g., https://en.wikipedia.org/wiki/Curvilinear_coordinates
Spherical coordinates are orthogonal curvilinear coordinates. Usually you use an orthonormalized basis at each point ("3-beins").
$$\vec{e}_r=\begin{pmatrix} \cos \varphi \sin \vartheta \\ \sin \varphi \sin \vartheta \\ \cos \vartheta \end{pmatrix}, \quad \vec{e}_{\vartheta} = \begin{pmatrix}\cos \varphi \cos \vartheta \\ \sin \varphi \cos \vartheta \\-\sin \vartheta \end{pmatrix}, \quad \vec{e}_{\varphi}=\begin{pmatrix}-\sin \varphi \\ \cos \varphi \\ 0 \end{pmatrix}.$$

dextercioby and Old Person
Old Person said:
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?
A long time ago, I wrote a German text that ... was the result of trying to make sense of such questions, but ended up somewhere completely different:
Es geht aber auch um drei sehr unterschiedliche Beweistechniken. Diese sind die Verwendung von Testkurven, die Manipulation von Integralen mit Testfunktionen, und die Begradigung eines Vektorfeldes.

Old Person said:
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?
At least that gives the formal definition of their meaning. When trying to work in different coordinates, looking at integrals over the expression in question applied to some test function is often the most intuitive way to make sense of the required mathematical manipulations. Often this approach works well, but sometimes other approaches (like relying on canonical transformation behavior) are more appropriate.

Old Person
gentzen said:
??? My impression is that the coordinate lines still intersect at right angles for spherical coordinates. Not that this would be important, but still, no need to make false statements.
Yes, I misspoke there.

Old Person and hutchphd
Old Person said:
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} ##
[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)) ##
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.
The important point when doing the calculations in curvilinear coordinates is that there also the basis vectors depend on the coordinates, and that this has to be taken into account when expressing the derivates in terms of these curvilinear coordinates. The most simple way is to use vector/tensor calculus and express the corresponding covariant derivatives like grad, div, and rot in the curvilinear coordinates, using their convariant definitions.

Here you can just calculate by "brute force". Expressed in their standard Cartesian components the spherical-coordinate unit vectors we need are
$$\vec{e}_{\vartheta}=\begin{pmatrix}\cos \varphi \cos \vartheta \\ \sin \varphi \cos \vartheta \\ -\sin \vartheta\end{pmatrix}, \quad \vec{e}_{\varphi}=\begin{pmatrix}-\sin \varphi \\ \cos \varphi \\ 0 \end{pmatrix}.$$
Now you can use your expression for ##\hat{\vec{L}}## (setting ##\hbar=1## for convenience), but you have to take into account that
$$\partial_{\vartheta} \vec{e}_{\vartheta}=-\vec{e}_r, \quad \partial_{\varphi} \vec{e}_{\vartheta}=\cos \vartheta \vec{e}_{\varphi}, \quad \partial_{\vartheta} \vec{e}_{\varphi}=0, \quad \partial_{\varphi} \vec{e}_{\varphi} = -\cos \vartheta \vec{e}_{\vartheta} - \sin \vartheta \vec{e}_r.$$
Using your expression you get [typos corrected in view of #11]:
$$\hat{\vec{L}}^2 \psi = \left (-\mathrm{i} \vec{e}_{\varphi} \partial_{\vartheta} + \mathrm{i} \vec{e}_{\vartheta} \frac{1}{\sin \vartheta} \partial_{\varphi} \right) \cdot \left (-\mathrm{i} \vec{e}_{\varphi} \partial_{\vartheta} \psi + \mathrm{i} \vec{e}_{\vartheta} \frac{1}{\sin \vartheta} \partial_{\varphi} \psi \right).$$
Now you take the derivatives, including those of the basis vectors and using that these form a Cartesian basis at any point, you get with pretty little effort the correct expression
$$\hat{\vec{L}}^2 \psi = -\frac{1}{\sin \vartheta} \partial_{\vartheta} (\sin \vartheta \partial_{\vartheta} \psi) - \frac{1}{\sin^2 \vartheta} (\partial_{\varphi}^2 \psi).$$

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TSny and Old Person
Thank you for all of the replies. DrClaude included - you've made less errors than I probably have and your general line of reasoning was sensible.

I agree with Gentzen, Nugatory and vanhees71, perhaps with a slight disgreement about the poles. There is some degeneracy at the poles but it is still possible to consider the vectors as being orthogonal. At the poles, the vector ##e_{\phi}## can be considered to become the 0 vector (so it cannot be normalised) but that will still produce an overall 0 dot product with any other vector and hence imply orthogonality. Similarly the x-component of the vector ##e_{\phi}## is not uniquely determined but that doesn't matter since the vector ##e_r## has 0 x-component there, so you will also produce an overall 0 dot product.

At the poles, of course spherical coordinates are singular.

Old Person
I've just seen the post added by vanhees71. Thank you.

This section:

## \hat{\vec{L}}^2 \psi = (-\mathrm{i} \vec{e}_{\varphi} \partial_{\vartheta} + \mathrm{i} \vec{e}_{\vartheta} \frac{1}{\sin \vartheta} \partial_{\varphi}) (-\mathrm{i} \vec{e}_{\varphi} \partial_{\vartheta} \psi + \mathrm{i} \vec{e}_{\vartheta} \frac{1}{2} \vartheta \partial_{\varphi} \psi). ##

does look like it has some small typing errors. There's a 1/2 on the right which should be 1/Sin ##\theta## for example - but I think I can see what you meant.

It looks like you have computed ##\hat{\vec{L}}^2## as the composition ## \hat{\vec{L}} ( \hat{\vec{L}} (\psi)) ## which I don't think is quite right. The composition would actually be impossible since the final output would have been a vector not a scalar and also the input for ##\hat{\vec{L}} ## cannot be a vector. For vector operators the expression ##\hat{\vec{L}}^2## must be understood as the dot product and not the composition.

However, I still think I can see the general point you are making. You are implying that the spherical unit vectors are acted on by the operators.

Thank you.

vanhees71
Can I also say thank you to whoever the moderators are. At some places (especially in the original post) they have fixed problems with equations produced with LaTeX that didn't display correctly. I added photos of what the equation should look like and they have obviously spent the time to fix it. Thank you.

I've corrected the typos. Of course the square of a vector (dot product of a vector with itself) is a scalar, not a vector. That's why I now put a dot between the brackets for clarity. Otherwise, I don't see what should be wrong with this derivation.

Old Person
Old Person said:
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? [...]
Short answer: it depends on the metric. In Cartesian coords, the metric is ##diag(1,1,1)##. In more general (curvilinear) coordinate systems the metric is nontrival, so the dot product is also more complicated.

First a warning: working with spherical coordinates contains a number of surprising booby traps. Often an "elementary" treatment uses normalized basis vectors, but if one tries to do things "properly" using an explicitly-covariant treatment (with metric and Christoffel symbols) one gets different answers, but only sometimes (sigh). I'll give a sketch of how to approach such computations using the latter method.

First off, if there a 3 ##L_i## operators in Cartesian coordinates, there must also be 3 in spherical coordinates. E.g., if we have ##L_z := x\partial_y - y\partial_x## (and cyclic),
a suitable triplet of operators in spherical coords takes the form:
$$R := \partial_\phi ~,~~~~ S := \cos\phi \;\partial_\theta ~,~~~~ T := -\sin\phi \;\partial_\theta ~-~ \cot\theta \, \cos\phi \;\partial_\phi ~.$$(See Carroll, sect. 5.2 for more. I'm using his notation. One can show that the ##R,S,T## satisfy the usual ##so(3)## commutation relations.)

To form the invariant square of these you need the spherical metric components: $$g_{ij} = diag(1, r^2, r^2 \sin^2\!\theta) ~,$$or rather the inverse components ##g^{ij}## which are the reciprocals component-wise. Here, of course, ##i,j,k## range over ##r,\theta,\phi##.

We'd like to write ##g^{ij} L_i L_j## but there's an ambiguity: should the metric be outside the two squares, or should it be in the middle, or... elsewhere? This dilemma is solved by replacing the partial derivatives by covariant derivatives in the formulas above. For this, you need to compute the Christoffel symbols corresponding to metric above. Using covariant derivatives means we can place the ##g^{ij}## out the front unambiguously, because the metric is covariantly-constant. (Such replacement of partial derivatives by covariant derivatives is easily seen to be legal because we could have done the same thing in Cartesian coords -- where the Christoffel symbols vanish.)

Then it's just a matter crunching out the many terms in ##g^{ij} L_i L_j##.

Good luck.

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Old Person, WernerQH, vanhees71 and 1 other person
Thanks strangerep.

I've got a copy of Spacetime and Geometry by Carroll so I will take another look at what you've said in more detail later. (I've already had a quick look and what you've said does make some sense).
It does seem to hinge around replacing the derivates in the operators ##\hat{L_x}## etc. with covariant derivatives and assuming that will be OK. I mean it should be, that is the entire premise of what is often called the "comma goes to semicolon" rule (page152 Carroll) - just assert that such a replacement is valid in any (possibly curved) space and with any co-ordinate system (without any hesitation or concern for proof).

More generally, I have managed to sort out the issue to my satisfaction anyway. You can resolve the issue from the expression given in post #8 by vanhees in a way that utilises nothing more than the definition of a dot product of vectors operators when they are written in cartesian form. Doing that does add a few lines and I won't bore everyone with the details.

Thank you to everyone who has spent some time here. I may not be following this thread any longer so please accept my apologies if I don't reply or put a "like" on any future comments.