# Magnitude of the cross product of two bra-kets?

From the mathematical perspective, one can take the magnitude of a cross product: $$|a\times b|=|a| |b| \sin{\theta}\cdot n,$$ where $$\theta$$ is the angle between a and b in the plane containing them, and n is the unit vector perpendicular to them.

Does this apply to the cross product of two bra-kets? I became curious from equation 10 on Dr. Berry's paper, where the Berry phase acquired by a state is the double integral of the following term: $$\langle \nabla n | m\rangle \times \langle m | \nabla n\rangle.$$

So, does the following make sense?

$$| \langle \nabla n | m\rangle \times \langle m | \nabla n\rangle| = | \langle \nabla n | m\rangle| | \langle m | \nabla n\rangle| \cdot\sin{\theta},$$

where

$$\cos{(\theta)}=\langle \nabla n | m\rangle \cdot\langle m | \nabla n\rangle.$$

To me, it doesn't make sense intuitively because cross products are usually given in tensorial notation in physics (for example: Expressing the magnitude of a cross product in indicial notation). Why is the application of the magnitude above incorrect, or when is it applicable?

• Note that the cross-product is defined in 3-dimensional space. What is the dimensionality of the Hilbert space of your quantum system? – dmckee Mar 9 at 23:03
• @Dan Yand Your comment looks like a fine tight answer to prevent misreadings of Berry, all right. – Cosmas Zachos Mar 9 at 23:52
• If @Dan's interpretation of the question is correct then my comment is way off base. :;sigh:: – dmckee Mar 9 at 23:54
• @dmckee Well, yes and no... It emphasizes that the Hilbert space as a vector space is aggressively irrelevant here... – Cosmas Zachos Mar 10 at 1:01
• Thank you, everyone. @DanYand it appears as if the mathematical definition for the magnitude of the cross product then holds. If someone could write that as an answer I will gladly accept it. As an aside, does this mean that we can use this convention to define the magnitude of Berry curvature? I am having trouble finding such magnitudes in the literature (except in numerical works). – TribalChief Mar 10 at 1:14

Usually, when we write $$|\cdots\rangle$$, the symbol(s) contained inside are just labels used to distinguish different elements of the Hilbert space from each other. That label can be a scalar, vector, matrix, or a list of our favorite arthropod species — but it's still labeling just a single element of the Hilbert space.

However, Berry's paper [1] is using a slightly unconventional notation $$|\nabla n\rangle$$ in which each component $$\nabla_k n$$ of $$\nabla n$$ labels a different element of the Hilbert space, so $$|\nabla n\rangle$$ is actually an abbreviation for three different kets, namely the kets $$|\nabla_k n\rangle$$ with $$k\in\{1,2,3\}$$.

Therefore, in Berry's equation ($$7$$c), the quantity $$v = \langle m|\nabla n\rangle$$ is a vector with three components $$v_k = \langle m|\nabla_k n\rangle$$ with $$k\in\{1,2,3\}$$. Its complex conjugate is $$v^*=\langle\nabla n|m\rangle$$. Both $$v$$ and $$v^*$$ are ordinary vectors with 3 complex components each. The fact that their components were constructed using bras and kets does not affect the meaning of the cross product $$v^*\times v$$.

The only unconventional feature of $$v^*\times v$$ is the fact that the components of the vectors are complex numbers, and part of the question is whether or not a familiar identity for the magnitude of the cross-product generalizes to this case. In particular, the relationship between $$|v^*\times v|$$ and $$v^*\cdot v$$ is questioned. To address this, write $$v=v_R+iv_I$$ where $$v_R$$ and $$v_I$$ are the real and imaginary parts of $$v$$. Then $$v^*\times v = 2i\,v_R\times v_I \hskip2cm v^*\cdot v = v_R\cdot v_R + v_I\cdot v_I.$$ This shows that the quantity $$v^*\times v$$ depends on the angle between $$v_R$$ and $$v_I$$, but the quantity $$v^*\cdot v$$ does not. Therefore, the value of $$v^*\cdot v$$ cannot determine the value of $$|v^*\times v|$$. This shows that the relationship questioned in the OP cannot hold in general.

As a check, consider the case $$v=(1,z,0)$$ with $$z=\exp(i\phi)$$. Then $$|v^*\times v|=2|\sin\phi| \hskip2cm v^*\cdot v=2$$ for all $$\phi$$. This confirms that $$v^*\cdot v$$ does not determine the value of $$|v^*\times v|$$.

Reference:

[1] Berry (1984), "Quantal phase factors accompanying adiabatic changes," https://michaelberryphysics.files.wordpress.com/2013/07/berry120.pdf

Actually, the entire equation is

$$\langle \nabla n |\times | \nabla n\rangle \rightarrow \sum_{m \neq n} \langle \nabla n | m\rangle \times \langle m | \nabla n\rangle$$

EDIT: It's Stoke's Theorem using a differential 2-form.

Stokes Theorem can be generalized to higher dimensions using differential forms.

The $$m\neq n$$ produces the wedge product.

And since $$\langle \nabla n | m\rangle$$ is the complex conjugate of $$\langle m | \nabla n\rangle$$ there are only 3 independent quantities.