Magnitude of the cross product of two bra-kets?

Question

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?

Answer 1

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 $|v^*\times v|$.

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Reference:

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

Answer 2

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.

See Differential Form