# How to understand this formula in quantum field theory?

In Condensed Matter Field Theory (page 288) by Altland and Simons, there is one mathematical formula:

$$$$\sum_\mathbf{p} \left(\mathbf{p}\boldsymbol{\cdot}\mathbf{v}\right)\left(\mathbf{p}\boldsymbol{\cdot}\mathbf{v}'\right)F\left(\mathbf{p}^2\right)\boldsymbol{=}\dfrac{\mathbf{v}\boldsymbol{\cdot}\mathbf{v}'}{d}\sum_\mathbf{p}\:\mathbf{p}^2F\left(\mathbf{p}^2\right) \tag{01}\label{01}$$$$

Here, $$\mathbf{p}$$ is the momentum, and $$\mathbf{v}$$ and $$\mathbf{v}'$$ are vectors. $$F\left(\mathbf{p}^2\right)$$ is a function of momentum which is invariant under rotational transformations; $$d$$ is the dimension considered (e.g., $$d=3$$ in $$x,y,z$$ coordinate system). My question is how to arrive at this formula?

Mathematically, this formula seems not correct. For example, imagine a summation over only $$\mathbf{p}\boldsymbol{=}\mathbf{p}_0$$ and imagine $$\mathbf{v}$$ and $$\mathbf{v}'$$ being orthogonal: then the left hand side is $$\left(\mathbf{p}_0\boldsymbol{\cdot}\mathbf{v}\right)\left(\mathbf{p}_0\boldsymbol{\cdot}\mathbf{v}'\right)F\left(\mathbf{p}_0^2\right)$$ which can be non zero, while the right hand side is zero because $$\left(\mathbf{v}\boldsymbol{\cdot}\mathbf{v}'\right)\boldsymbol{=}0$$. I am wondering if there are some arguments in physics to "prove" this formula. For example, if the summation is replaced by an integration over the whole reciprocal space of momentum p with d dimension, will the formula stand?

• It certainly isn't true without the sum over all ${\bf p}$, but that's neither here nor there.
– Buzz
Jan 21 '21 at 2:08
• The summation is likely over all p. Jan 21 '21 at 2:09
• If the summation is over all p, how to prove such a formula? Thanks. Jan 21 '21 at 2:22
• @Solidstate presumably starting by using the definitions of the dot product being $a\cdot b= a_{\mu}b_{\nu} g^{\mu \nu}$ and so on, though I haven't given it much thought yet. Jan 21 '21 at 2:36
• Thank you so much! Jan 21 '21 at 14:05

The formula holds, because there is a sum over all $${\bf p}$$. After that sum is performed, $${\bf v}\cdot{\bf v}'$$ is essentially the only vector structure that is consistent with rotation symmetry.

To prove the formula, first rewrite it so that the $${\bf v}$$ and $${\bf v}$$' are outside the sum: $$\sum_{{\bf p}}({\bf p}\cdot{\bf v})({\bf p}\cdot{\bf v}')F(p^{2})= \sum_{j=1}^{d}v_{j}\sum_{j=1}^{d}v_{k}'\sum_{{\bf p}}p_{j}p_{k}F(p^{2}).$$

Now look just at the final sum, $$\sum_{{\bf p}}p_{j}p_{k}F(p^{2})$$. Consider holding the index $$j$$ fixed (say $$j=1$$) and varying $$k$$. If $$k\neq j$$ (say $$j=2$$) then for every term in the sum with $${\bf p}=(p_{j},p_{k},p_{3})$$, there is another term in the sum with $${\bf p}'=(p_{j},-p_{k},p_{3})$$. Since the sum terms $$p_{j}p_{k}F(p^{2})$$ and $$p_{j}'p_{k}'F(p'^{2})$$ are equal in magnitude but opposite in sign, they will cancel; and, if fact, every term in the sum will be canceled by a similar one with an inverted $$p_{k}\rightarrow p_{k}'=-p_{k}$$. Thus, the sum is $$\sum_{{\bf p}}p_{j}p_{k}F(p^{2})=0$$ if $$j\neq k$$.

The expression $$\sum_{{\bf p}}p_{j}p_{k}F(p^{2})$$ transforms like a two-index tensor. (Any product of the components of two vectors $${\bf u}$$ and $${\bf w}$$ along with a scalar $$\phi$$—that is, $$u_{j}w_{k}\phi$$—transforms this way. The sum of interest is then just a sum of tensors that transform this way and so transforms this way itself.) The only two-tensor that has components that are nonzero only for $$j=k$$ is one that is proportional to the identity, with components $$\delta_{jk}$$. [This is essentially just a long-winded way of saying that, since there is no preferred direction, $$\sum_{{\bf p}}p_{1}p_{1}F(p^{2}) =\sum_{{\bf p}}p_{2}p_{2}F(p^{2})=\sum_{{\bf p}}p_{3}p_{3}F(p^{2})$$.] So we must have $$\sum_{{\bf p}}p_{j}p_{k}F(p^{2})$$ be a scalar function times $$\delta_{jk}$$, $$\sum_{{\bf p}}p_{j}p_{k}F(p^{2})=\delta_{jk}G,$$ which leaves us to find $$G$$.

To get the scalar $$G$$, we contract the tensor with another $$\delta_{jk}$$, or $$\sum_{j=1}^{d}\sum_{k=1}^{d}\delta_{jk}\sum_{{\bf p}}p_{j}p_{k}F(p^{2})=\sum_{j=1}^{d}\sum_{k=1}^{d}\delta_{jk}\delta_{jk}G,$$ which reduces to $$\sum_{{\bf p}}p^{2}F(p^{2})=dG,$$ since $$\sum_{j=1}^{d}\sum_{k=1}^{d}\delta_{jk}\delta_{jk}$$ is just the dimensionality of space $$d$$.

Then it is just a matter of inserting this expression into the earlier ones and working backwards. Solving for $$G$$, we have $$\sum_{{\bf p}}p_{j}p_{k}F(p^{2})=\delta_{jk}G=\delta_{jk}\frac{1}{d}\sum_{{\bf p}}p^{2}F(p^{2}).$$ Then inserting this into the original expression with $$v_{j}$$ and $$v_{k}'$$ gives $$\sum_{j=1}^{d}v_{j}\sum_{j=1}^{d}v_{k}'\sum_{{\bf p}}p_{j}p_{k}F(p^{2})= \sum_{j=1}^{d}v_{j}\sum_{j=1}^{d}v_{k}'\delta_{jk}\frac{1}{d}\sum_{{\bf p}}p^{2}F(p^{2}),$$ and collecting this in component-free form gives the final expression $$\sum_{{\bf p}}({\bf p}\cdot{\bf v})({\bf p}\cdot{\bf v}')F(p^{2})=\frac{1}{d}{\bf v}\cdot{\bf v}'\sum_{{\bf p}}p^{2}F(p^{2}).$$

• Interesting answer, although it's much easier to make use of the identity $p_{\mu}p_{\nu} = \frac{1}{d} g_{\mu \nu} p^2$, whose proof is i believe similar if not shorter than what you did. Jan 21 '21 at 3:26
• Thank you so much. Very brilliant idea! Jan 21 '21 at 13:33