I'm hoping someone might be willing to help me obtain an understanding of the differential forms notation ubiquitous in QFT and GR. My problem is I'm fairly new to tensor algebra/calculus but I don't know the best way to obtain the background I need to go forward. I understand that $\Box\equiv\partial^{\mu}\partial_{\mu}$ is just the spacetime derivative in Einstein summation and with the Minkowski metric $\eta^{\mu\nu}$. My biggest question is what is the difference between superscripts and subscripts and is there any significance to when $\nu$ is used as opposed to $\mu$ or vice versa? How do I best understand things like $\partial^{\mu}\phi\partial_{\mu}\phi$ in the free part of the QFT Lagrangian? How do I understand the subscript and superscript $\mu$'s and $\nu$'s in the electromagnetic tensor in its derivative form $F_{\mu\nu}\equiv\partial_{\nu}A_{\mu}-\partial_{\mu}A_{\nu}$? Are these notations somewhat arbitrary, or is there a relevant physical meaning to the exact way the symbols and their placements are arranged?

  • 4
    $\begingroup$ Physics students usually first encounter this notation in a graduate-level course on electromagnetism and Special Relativity, before studying QFT. $\endgroup$
    – Ghoster
    Nov 9, 2022 at 3:01
  • 1
    $\begingroup$ This is index notation, which tracks components of a tensor (or vector or...)...though I guess wikipedia calls it Ricci calculus for index gymnastics. $\endgroup$ Nov 9, 2022 at 3:03
  • $\begingroup$ I would recommend Weinberg's excellent book on GR, the chapters on Special Relativity. $\endgroup$ Nov 9, 2022 at 3:06
  • $\begingroup$ See physics.stackexchange.com/q/735271 $\endgroup$
    – Ghoster
    Nov 9, 2022 at 3:11
  • 1
    $\begingroup$ @Ghoster physics students definitely encounter this notation in undergrad courses nowadays $\endgroup$ Nov 9, 2022 at 8:42

2 Answers 2


There's a lot behind all of this but if you're already in a QFT course I think it's not the right time to take a long time to deeply understand it. You can spend months on this. But you would do well to know how to use it, and that should be sufficient for what you need for now. I'll just give you a few tidbits that can serve you in your course.

  1. There is no reason to use $\mu$ rather than $\nu$ or $\sigma$ or $\rho$. But you should be able to identify, in an equation with these indices, which indices are free indices and which are summed. For example, in

$$F_{\mu \nu} = \partial_\nu A_\mu - \partial_\mu A_\nu$$

since by Einstein notation there are no sums these are free. Free indices means that, next to the equation, there is an imaginary "$\forall \,\mu, \nu$" meaning that the equation holds for every possible combination of the pair $\mu, \nu$ equal to 0,1,2, or 3. That means the equation above is actually shorthand for 16 different equations. Any time greek letters like $\mu, \nu, \rho, \sigma, \alpha, \beta, \delta, \gamma$ are used as indices, it's usually implied that the sum goes over 0,1,2,3. If the indices are $i,j,k$ the sum goes from 1 to 3.

An example of an expression without any free indices is

$$\partial^\mu \phi \partial_\mu \phi$$

and here there is no implicit "for all" statement... this quantity is just summed and no variable indices are left over.

SANITY CHECK FOR FREE INDICES: Make sure that your equations are correct by checking that there is an equal number of free indices in every term, and they are the same index letters. For example if $\mu$ is free, there should be exactly one upper-index $\mu$ in every term of the equation. If not you did something wrong.

SANITY CHECK FOR SUMMED INDICES: There must be one upper (contravariant) index for every lower (covariant) index in a sum. Why? We organize our equations this way because such quantities do not change form under transformation, whereas if we sum together two of the same type of index the transformations get messy. An index can be raised or lowered using the "metric", which in special relativity is written $\eta ^{\mu\nu}$, or $\eta _{\mu \nu}$, depending which one you need (raising or lowering):

$$A^\mu = \eta^{\mu \nu} A_\nu$$ $$A_\mu = \eta_{\mu \nu} A^\nu$$ 2. What is the difference between upper and lower indices?

There's a whole intro I could give here to motivate these indices but you need information fast. So really it's just that they remind you how that quantity changes when you change reference frame through a rotation or a boost. For each upper index (called a contravariant index), you multiply by the matrix with components $\Lambda ^\mu _{\,\,\,\nu}$:

$$A'^{\mu} = \Lambda ^\mu _{\,\,\,\nu} A^\nu $$

Notice that I left space under the $\mu$ index in $\Lambda$ above. That is because if you just put the indices of a matrix on top of each other, you lose track of which represents the column and which is the row. And we know that for a matrix, $M_{ij} \neq M_{ji}$ necessarily. Here, $\mu$ is free and $\nu$ is summed over. If the first index is rows and the second is columns (this is always the convention) then the second index is summed over. I remember it by thinking that the summed index $\nu$ wants to get as close as it can to its summed "partner", the other $\nu$ .

So, that's how the superscripts transform when changing frame. And if you have two superscripts you need to just multiply by two of those matrices. Just make sure that at the end, you have two free indices and they are the same on both sides of the equation. For the summed indices you can invent any index you want:

$$T'^{\mu \nu} = \Lambda^\mu _{\,\,\, \rho} \Lambda ^\nu_{\,\,\, \sigma} T^{\rho \sigma}$$

For example above I was careful that $\mu, \nu$ are matching on both sides, but I made up the indices $\rho$ and $\sigma$ here just to write out the sum, just like you know that

$$1 + 2 + 3 = \sum_{n=1,2,3} n = \sum_{m=1,2,3} m = 1 + 2 +3$$

...so the index that I use to write the sum gives the same result no matter what.

That covers superscripts. For every superscript your quantity transforms as above. Subscripts (called covariant indices) are the same but you multiply by a matrix with components $\Lambda _{\mu} ^{\,\,\, \nu}$ instead. This seems awfully similar, but if you plug in values for $\mu$ and $\nu$, you will see that these matrices have different numbers. However, you usually don't need to do that, and we are just doing the essentials here, so I'm not going to write the explicit form of the Lorentz transformation matrix. Here is an example of the transformation of a covariant object:

$$A'_{\mu} = \Lambda _\mu ^{\,\,\,\nu} A_\nu $$

And finally, you should remember (without proof here) that any contracted index of one upper + one lower index doesn't change form at all in the transformation. The matrices are each other's inverse, and they cancel out, so that

$$A'_\mu A'^\mu = A_\mu A^\mu $$

This property makes for a lot of convenience. Note that even though $A^\mu$ is "a vector", once you write an index these are just the components here which are numbers, so the factors would commute in the equation above if you want them to. Ordinary numbers commute.

Note, why do the $\Lambda$ matrices need one upper and one lower index? Let's say we are transforming a contravariant (upper) index. We will have a sum, and that sum has to be with a lower index, so one of the indices of $\Lambda$ must be lower. But in the end, we need to end up with an upper index on both sides, so that we fulfill the sanity check for free indices. So we must also have an upper index, and so $\Lambda$ has one upper and one lower.

How do I best understand things like $\partial^{\mu}\phi\partial_{\mu}\phi$ in the free part of the QFT Lagrangian?

From Einstein notation (which I guess you are familiar with) this is just

$$\partial^{0}\phi\partial_{0}\phi + \partial^{1}\phi\partial_{1}\phi + \partial^{2}\phi\partial_{2}\phi + \partial^{3}\phi\partial_{3}\phi $$

where the 0,1,2,3 components in most notations stand for $t,x,y,z$. Note that there is one upper index and one lower index being summed.

How do I understand the subscript and superscript $\mu$'s and $\nu$'s in the electromagnetic tensor in its derivative form $F_{\mu\nu}\equiv\partial_{\nu}A_{\mu}-\partial_{\mu}A_{\nu}$? Are these notations somewhat arbitrary, or is there a relevant physical meaning to the exact way the symbols and their placements are arranged?

There are two free indices on both sides of the equation. So there is an implicit "for all $\mu, \nu$" meaning this is actually 4 x 4 = 16 equations:

$$F_{00}\equiv\partial_{0}A_{0}-\partial_{0}A_{0}$$ $$F_{01}\equiv\partial_{1}A_{0}-\partial_{0}A_{1}$$ $$F_{10}\equiv\partial_{0}A_{1}-\partial_{1}A_{0}$$ $$...$$

Last thing: If you start reading about differential forms, considering you are already in QFT, I would recommend staying away from that. Mathematics has a whole parallel way of talking about these objects and while you can benefit from that in the long term if you want to go very deep, in the short term it is only a huge source of confusion. Stay away for the time being from differential forms until you have the time and peace of mind to learn them properly (if you wish to). You don't need that language for an introductory QFT course.

  • $\begingroup$ I think you explained it all except Tensors' definition as a multilinear map (which is implicit here). What advanced topics did you have in mind (just the names)? $\endgroup$
    – Ryder Rude
    Nov 9, 2022 at 8:57
  • 1
    $\begingroup$ There are nice ways of motivating why you even need two different indices and where everything comes from - first of all in non-orthonormal bases, and then extending to the metric in SR, and talking about the metric and eventually also the christoffel symbols. The stuff I wrote in this post is a bit "magic" still, in my opinion, but I think it's what OP needs since he's already 4 steps ahead, in QFT. $\endgroup$ Nov 9, 2022 at 12:54

Superscript vs Subscript Indices

I don't understand QFT that well but I have a decent understanding of GR and tensor calc. In the case of GR, subscript and superscript indices signify covariance and contravariance respectively which are linear transformation concepts. In a vector space, tensor components (which include vectors) are considered covariant if they change using the same linear transformation as basis vectors. This can be thought of as tensor components also increasing when basis vectors increase under a change in unit length. For example: a 1-form ω (aka a covector), which is expanded the following way in a vector space V, $$\omega=\omega_\mu\epsilon^\mu$$ contain components which can be interpreted as the 1-form acting on a corresponding basis vector in the vector space. $$\omega_\mu=\omega(\overrightarrow{e_\mu})$$ enter image description here

In this case, our 1-form components change according to the linear transformation: $$\tilde\omega_\nu=\Sigma_{\nu=0}^nA_{\mu\nu}\omega_\mu$$ where I'll denote the matrix A as the covariant linear transformation matrix whose components are the coefficients that relate ω_ν to ω_μ just as the corresponding linear transformation applied to basis vectors also satisfies the matrix A: $$\overrightarrow{e_\nu}=\Sigma_{\nu=0}^n A_{\mu\nu}\overrightarrow{e_\mu}$$ On the other hand, contravariant vector components are ones that change in the opposite way as basis vectors. For example: decreasing the magnitude of a basis vector increases the value of a vector's components. $$\overrightarrow{v}=v^\mu\overrightarrow{e_\mu}$$ Here, the matrix A used to transform basis vectors and 1-forms no longer works with vector components. Therefore, a new contravariant linear transformation matrix B is required so that $$v^\nu=\Sigma_{\nu=0}^n B_{\mu\nu}v^\mu$$ where within the same vector space V, the coefficients in B will either be half or twice the value of the coefficients in A depending on whether A's coefficients increase or decrease respectively.

Differentiating between μ and ν just means that μ and ν are not necessarily of equal value (with the exception of a symmetric matrix which, by definition, means that μ = ν).

To learn more, I recommend watching Eigenchris' playlist on tensor calculus or the book Introduction to Tensor Analysis and the Calculus of Moving Surfaces by Pavel Grinfeld

D'alembert Operator Acting On A Function

Additionally, the d'alembert operator acting on a scalar function Φ can be expanded to reproduce the wave equation (written using the (+,-,-,-) sign convention): $$c^2\frac{\partial^2 \phi}{\partial x^2}=\frac{\partial^2 \phi}{\partial t^2}$$ $$\Box \phi=\partial^\mu\partial_\mu \phi=\eta^{\mu\nu}\partial_\mu\partial_\nu \phi=0$$ $$\partial_0^2\phi-\partial_1^2\phi-\partial_2^2\phi-\partial_3^2\phi=0$$ $$\frac{1}{c^2}\frac{\partial^2 \phi}{\partial t^2}-\frac{\partial^2 \phi}{\partial x^2}-\frac{\partial^2 \phi}{\partial y^2}-\frac{\partial^2 \phi}{\partial z^2}=0$$ $$c^2\left(\frac{\partial^2 \phi}{\partial x^2}+\frac{\partial^2 \phi}{\partial y^2}+\frac{\partial^2 \phi}{\partial z^2}\right)=\frac{\partial^2 \phi}{\partial t^2}$$

Note: tensor indices written in greek characters represent the consideration of time as a dimension whereas indices written in Latin characters discount time, only referring to spatial dimensions.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.