In this paper, the author discusses the following nonlinear response relation: $$\delta s_a=\alpha_{abc}E_bE_c \tag{1},$$ where $\delta s_a$ represents the spin polarization generated by the current driven by the electric field $E_b/E_c$. In Eq.(1), $E_b/E_c$ are time-reversal-invariant quantities: if $\delta s_a$ is reversed by time-reversal symmetry ($\mathcal{T}$), we conclude that $\alpha_{abc}$ is a T-odd tensor; if $\delta s_a$ is not reversed by time-reversal symmetry, we conclude that $\alpha_{abc}$ is a T-even quantity.

However, in the paper cited before, the author claims $$\alpha_{abc}=\alpha_{abc}^{even}+\alpha_{abc}^{odd} \tag{2}$$ I don't understand why this partition is legitimate. Is it due to the sign change of $\delta s_{a}$ under time-reversal symmetry being indefinite?

Furthermore, consider the nonlinear current response: $$j_a=\sigma_{abc}E_bE_c \tag{3}$$ where the current density $j_a$ will change its sign under time-reversal symmetry and the electric field $E_b/E_c$ will remain unchanged under time-reversal symmetry. As a result, we find: $$\mathcal{T}\sigma_{abc}=-\sigma_{abc}$$ namely, the nonlinear conductivity should be a T-odd tensor. However, in another paper, the author seems to also admit a similar partition: $$\sigma_{abc}=\sigma_{abc}^{even}+\sigma_{abc}^{odd}$$

How could this be true? The current density is not reversed by time-reversal symmetry? This can not be the case.


1 Answer 1


I don't think this is a statement about the particular tensors $\alpha$ and $\sigma$, this is a general result.

Tensors as sums of even and odd terms

Suppose we have a tensor $M$, and an operator $\mathcal{T}$ with eigenvalues $\pm 1$ (an idempotent operator, e.g. a parity operator). We can always write:

$$ \begin{align} M &= M + \frac{1}{2}\left(\mathcal{T}M - \mathcal{T}M\right)\\ &= \frac{1}{2}\left[(M + \mathcal{T}M) + (M - \mathcal{T}M)\right] \tag 1 \end{align} $$

Consider now the first term, $(M + \mathcal{T}M)$; if we apply $\mathcal{T}$ to it, we get:

$$ \begin{align} \mathcal{T}(M + \mathcal{T}M) &= (\mathcal{T}M + \mathcal{T}^2M) \\ &= \mathcal{T}M +M \\ &= \left(M + \mathcal{T}M\right) \tag 2 \end{align} $$ where we have used the fact that $\mathcal{T}^2=\mathrm{I}$ (the identity), since it is idempotent; i.e. its eigenvalues are all +1 or -1, so the squared eigenvalues are all +1.

Equation $(2)$ shows that $(M + \mathcal{T}M)$ is unchanged when we apply $\mathcal{T}$, so this first term in equation $(1)$ is $\mathcal{T}$-even.

Now we repeat this for the second term in equation $(1)$,

$$ \begin{align} \mathcal{T}(M - \mathcal{T}M) &= (\mathcal{T}M - \mathcal{T}^2M) \\ &= \mathcal{T}M -M\\ &= -\left(M - \mathcal{T}M\right). \tag 3 \end{align} $$ Now we see that applying $\mathcal{T}$ has resulted in minus what we started with, and therefore this term is $\mathcal{T}$-odd.

Thus, equation $(1)$ expresses the tensor $M$ as a sum of a $\mathcal{T}$-even and a $\mathcal{T}$-odd tensor. At no point have we assumed anything about our tensor $M$, and so this expression holds generally. $$ \begin{align} M &= \frac{1}{2}\left(M + \mathcal{T}M\right) + \frac{1}{2}\left(M - \mathcal{T}M)\right)\\ &= M^\mathrm{even} + M^\mathrm{odd} \tag 4 \end{align} $$

When $M$ is $\mathcal{T}$-even or odd

If the tensor $M$ does have a particular $\mathcal{T}$-symmetry, then all that happens is that one of the terms in equation $(4)$ is zero (i.e. either $M^\mathrm{even}=0$ or $M^\mathrm{odd}=0$). For example, if $M$ is $\mathcal{T}$-odd, then $\mathcal{T}M=-M$ and we have:

$$ \begin{align} M^\mathrm{even} &= \frac{1}{2}\left(M + \mathcal{T}M\right) \\ &= \frac{1}{2}\left(M + (-M)\right) \\ &= 0; \end{align} $$ and, correspondingly, $$ \begin{align} M^\mathrm{odd} &= \frac{1}{2}\left(M - \mathcal{T}M\right) \\ &= \frac{1}{2}\left(M - (-M)\right) \\ &= \frac{1}{2}\left(2M\right) \\ &= M, \end{align} $$ which is as expected, since in this example $M$ is $\mathcal{T}$-odd!

  • $\begingroup$ Many thanks for your excellent answer. The parity operator $\mathcal{P}$ works for your partition in Eq.(1), what about the time-reversal operator? I mean $\mathcal{T}^2=-1$ will be obtained for time-reversal symmetry. In addition, is there any reference for the partition in Eq.(1)? $\endgroup$
    – Jack
    Oct 4, 2022 at 6:17
  • $\begingroup$ @Jack to get equation 1, you just add and subtract TM from M; I've edited the answer to show this. For complex Hermitian operators (e.g. time-reversal), I think it's actually T-dagger T you do, so that you still get 1 (|T|^2=1). $\endgroup$ Oct 4, 2022 at 12:38
  • $\begingroup$ The first paper has the quote: "Note that in nonmagnetic materials, only $\alpha^\text{even}$ exists. The two parts have different symmetry properties. For $\alpha^\text{odd}$ these can be found in [32]." So at least in this paper, the partition to odd/even isn't just a mathematical trick, but seems to convey a real physical meaning. $\endgroup$
    – Tyberius
    Oct 4, 2022 at 12:49
  • $\begingroup$ @Tyberius It often does have a physical meaning, but I thought the question was about how such a partitioning is possible without assuming any particular properties. $\endgroup$ Oct 4, 2022 at 12:55
  • $\begingroup$ @PhilHasnip Hi, bother you again. I understand the partition trick, one is even and the other one is odd. The proof depends on $\mathcal{T}^2=1$, this works for parity symmetry but is not true for the time-reversal operator, which demands $\mathcal{T}^2=-1$, seeing Eq.(14) its.caltech.edu/~xcchen/img/Ph129b2020/lecture/lecture0312.pdf $\endgroup$
    – Jack
    Oct 5, 2022 at 0:33

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