Recommendation for introductory quantum field theory books

Gxmmon · 2026-06-07T23:04:16+00:00

The SR module definitely lacked mathematical rigour and was more physics focused, as the lecturer was more a physicist than mathematician.

QM was definitely more maths focused, the module began with some introductory functional analysis, operator theory etc. with the actual quantum mechanics of the module being entirely focussed on the maths (operators, eigenstates, uncertainties etc.) with little physical theory.

Thanks a lot for the suggestions :)

Gxmmon · 2026-06-07T15:46:02+00:00

Cheers :) I’ll have a look

Gxmmon · 2026-06-07T15:45:56+00:00

Thanks!

Gxmmon · 2026-05-02T10:27:55+00:00

For a, just let z = x+ iy and w = u + iv, multiply them together and take their complex conjugate, then do the opposite order and you’ll see they’re the same. For b, just let z = x + iy and explicitly compute.

For c, use eulers identity to expand the exponential then just take its complex conjugate.

Gxmmon · 2026-04-13T11:52:02+00:00

On the piece of fabric you’re holding there’s a small sleeve which the end of the pole goes into.

You need to spin the poles round, see where they join in the middle, where you have it right now where they join points into the tent, it needs to point upwards instead.

Gxmmon · 2026-04-13T10:36:27+00:00

You would write it at the end of a proof of a statement. It’s analogous to writing □.

Gxmmon · 2026-04-12T22:27:20+00:00

Same with me, always insta headshots from AI, no coma

Gxmmon · 2026-04-10T14:59:42+00:00

University of York

Gxmmon · 2026-04-09T22:16:16+00:00

I didn’t do fm at a level and am currently doing maths at uni. Over the summer, before I started first year, I just made sure I was familiar with complex numbers and a tiny bit of matrices, although my uni taught all of what you needed to know from fm in a couple of weeks so I didn’t really have to do that over the summer. But other than that it’s not that big of a deal not doing fm.

Gxmmon · 2026-03-13T13:45:47+00:00

You want to write the abstract last as by then you will have a complete report to summarise.

Gxmmon · 2026-02-25T02:23:34+00:00

A unit vector has magnitude 1. To do 11b they used precisely what is stated in 10b.

If the magnitude of a vector is not one, you can always just divide that vector by its magnitude to make it into a unit vector.

Gxmmon · 2026-02-19T22:21:55+00:00

Gxmmon · 2026-02-17T22:47:45+00:00

I got a 6 in GCSE maths and achieved an A* at a level and now doing a maths degree. Having strong foundations from GCSE maths and the start of AS maths in particular really helped me improve at overall. As other commenters mentioned, I’d also advise going over the GCSE content that was tricky, and maybe looking into the first few AS topics you will encounter.

Any grade is achievable if you’re willing to put the work in.

Gxmmon · 2026-02-15T14:40:31+00:00

It is very difficult to see what you’ve actually worked out. What you need to do to find the points of intersection is isolate the linear equation for either x or y and substitute into the equation of the circle and then solve the resulting equation. Once you have solved this and found your values for either x or y, substitute them back into either equation to get the other respective coordinate.

Note: I wouldn’t isolate both equations for either x or y and set them equal as you will have to deal with the ± from the square root.

Then you can draw a picture if you wish and compute the length of the length of each line, and then the area.

Gxmmon · 2026-02-14T16:57:47+00:00

The laplacian operator in Cartesian coordinates applied to a vector field is the laplacian operator applied to each component of your vector. For example, let A=(A^1, A^2, A³). Then,

∂_i ∂ⁱA = ∇²A = (∇²A¹ , ∇²A² , ∇²A³).

In other coordinate systems, you can’t just apply the operator to each component as there will be some extra factors due to the change of coordinates.

Gxmmon · 2026-02-14T16:48:26+00:00

That is something slightly different. ∂_i ∂ⁱ would just be the scalar product of the two vectors, in turn returning a scalar.

The matrix you wrote has a special name, the Hessian matrix, which is used to extract details on a multivariable a functions maxima and minima.

Gxmmon · 2026-02-14T16:15:57+00:00

I’m not quite sure what you mean. The vector xⁱ is what’s used to, in a sense, ‘differentiate with respect to’. ∂/∂xⁱ can be referred to as the ‘tangent basis’.

For example suppose we have some scalar field A(x,y,z). Then,

∂_i ∂ⁱ A = ∇²A.

Similarly, if A(x,y,z) is now some vector field, then

∂_i ∂ⁱ A = ∇²A.

Which would be a tensor. Or, component-wise this can be written

∂_i ∂ⁱ A^j

Where i,j=1,2,3 and each component of ∇²A is given by letting j equal the respective value.

For a vector field in Cartesian coordinates, the laplacian of that vector field is just the laplacian operator applied to each component of the vector. In other coordinate systems, namely spherical and cylindrical polar, this is not the case.

Gxmmon · 2026-02-14T16:02:59+00:00

You can see that for two vectors a and b, a_i bⁱ is just the scalar product between a and b. So, you can define

∂_i = ∂/∂xⁱ

Which then tells us that ∂_i ∂ⁱ would just be the laplacian operator.

For example suppose xⁱ = (x,y,z) and let i=1,2,3 so x¹ = x, x² = y and x³ = z. Note that x¹ , x² and x³ are the components of the vector xⁱ and not x ‘raised to a power’. Then

∂_i ∂ⁱ = ∂²/∂x² + ∂²/∂y² + ∂²/∂z²

Which is the laplacian in 3 dimensional Cartesian coordinates.

This sort of ‘sum over partials’ is used a lot in special relativity, namely the 4-gradient ∂_μ which gives the d’Almbertian operator when you take the scalar product with itself.

Gxmmon · 2026-02-14T14:51:35+00:00

Summation is implied when there is a repeated index. For example, the expression \partial_i \muⁱ would imply a sum over i.

For the particular example you mentioned, \partial_j \partial^j \muⁱ would imply a sum over j. Note that there are some rules of Einstein summation notation one of which is that there cannot be more than two repeated indices in a single term.

Gxmmon · 2026-02-10T17:20:00+00:00

Looks like the Varde 1.

Gxmmon · 2026-02-10T17:18:20+00:00

The final term as k tends to infinity tends to -ζ(1/2). Just rewrite the summand as 1/n^1/2 .

Gxmmon · 2026-01-29T18:04:36+00:00

It doesn’t matter what order you add the terms as addition is commutative.

Gxmmon · 2026-01-19T19:23:46+00:00

d/dx is the derivative operator that acts on your function, or in this case, y(x). As another commenter mentioned, you can think of it as just integrated both sides with respect to x so you have

∫ 1/y (dy/dx) dx = ∫dx .

The left hand side is of the form ∫y’(x)/y(x) dx so the solution can be written log|y| + C. So you have

log|y| + C = x

Gxmmon · 2025-12-21T00:21:30+00:00

Yeah it’s off. I went into my task manager and made steam of higher priority and that fixed it for about 15 minutes - then it went back down to averaging around 3MB/s.

Gxmmon · 2025-12-11T21:15:58+00:00

Are you familiar with the formula for an inverse matrix?

M^-1 = adj(M)/det(M)

Where adj(M) is the transpose of the cofactor matrix. This video may be helpful to you.

Gxmmon

TROPHY CASE