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

\begin{equation} \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} \end{equation}

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?

  • $\begingroup$ It certainly isn't true without the sum over all ${\bf p}$, but that's neither here nor there. $\endgroup$
    – Buzz
    Jan 21 '21 at 2:08
  • $\begingroup$ The summation is likely over all p. $\endgroup$
    – my2cts
    Jan 21 '21 at 2:09
  • $\begingroup$ If the summation is over all p, how to prove such a formula? Thanks. $\endgroup$
    – Solidstate
    Jan 21 '21 at 2:22
  • $\begingroup$ @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. $\endgroup$
    – Triatticus
    Jan 21 '21 at 2:36
  • $\begingroup$ Thank you so much! $\endgroup$
    – Solidstate
    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 $$\sum_{{\bf p}}({\bf p}\cdot{\bf v})({\bf p}\cdot{\bf v}')F(p^{2})=\frac{1}{d}{\bf v}\cdot{\bf v}'\sum_{{\bf p}}p^{2}F(p^{2}).$$

  • $\begingroup$ 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. $\endgroup$
    – Triatticus
    Jan 21 '21 at 3:26
  • $\begingroup$ Thank you so much. Very brilliant idea! $\endgroup$
    – Solidstate
    Jan 21 '21 at 13:33

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