# Bras and Kets and Tensors

Hurkyl
Staff Emeritus
Gold Member
mrandersdk-

If |00>,|01>,|10> and |11> (1=up,0=down)

are linear independent vectors, then <01|01> = 0,

rather than <01|01> = <0|0><1|1>, as you suggest.
How do you figure?

You may have an argument in that I implicitly assume that in $R\otimes R$ one is a row vector and the other is a column vector, so an nx1 vector times a 1xn vector is an nxn matrix, but I wouldn't even know how to express a transpose operation at higher ranks without people loosing track of the otherwise very
simple math.
Regards, Hans
Transposition is more of a notational device, than anything, to keep track of where the rows and columns are.

In higher ranks, you can use labels to keep track rows, columns, depth..., and use a modified Einstein summation to multiply matrices.

$$Y = M^{T} \Rightarrow Y_{cr} = M_{rc}$$

$$(M_{abc...z} N_{abc...z})_{(fg)} = \stackrel{\Sum (M_{abc...z} N_{abc...z})}{f,g=i, i=1...n}, f$$
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Any mistakes I blame on LaTex

You may have an argument in that I implicitly assume that in $R\otimes R$ one is a row vector and the other is a column vector, so an nx1 vector times a 1xn vector is an nxn matrix, but I wouldn't even know how to express a transpose operation at higher ranks without people loosing track of the otherwise very
simple math.

Regards, Hans
Transposition is more of a notational device, than anything, to keep track of where the rows and columns are. Which elements combine with which elements between two tensors is unchange by

In higher ranks, you can use labels to keep track of rows, columns, depth...etc, and use a modified Einstein summation to multiply matrices.

$$Y = M^{T} \Rightarrow Y_{cr} = M_{rc}$$

$$(M_{abc...f} N_{c\: d\: e...z})_{(dp)} \equiv \sum_{d_i , p_i \ i=1...n} (M_{abc...f} N_{c\: d\: e...z})}\ , \ \ \ \ d \neq p$$

$$L_{abc_{m}e_{m}f_{m}c_{n}e_{n}f_{n}ghi...o,qrs...z} = (M_{abcef} N_{efg...z})$$

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Any mistakes now, in the past, or ever, I blame on LaTex, whether I'm using it or not.

mrandersdk-

If |00>,|01>,|10> and |11> (1=up,0=down)

are linear independent vectors, then <01|01> = 0,

rather than <01|01> = <0|0><1|1>, as you suggest.
no, $$|01>^\dagger = <01|$$

mrandersdk, Hurkl-

I posted:
If |00>,|01>,|10> and |11> (1=up,0=down)

are linear independent vectors, then <01|01> = 0,

rather than <01|01> = <0|0><1|1>, as you suggest.

How do you figure?
I figure, I misread <01|01> as <01|10>

(I wouldn't mind if someone deleted my extra and partially edited post, #152.)