How Do Ket Vectors Change When Working With Bra Tensor Products?

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SUMMARY

The discussion clarifies the transformation of ket vectors to bra vectors when applying tensor products in quantum mechanics. Specifically, when working with the tensor product of two states |v1>|v2>, the corresponding bra representation is |v2> is calculated using the bra representation |v2>. This notation emphasizes the correct application of tensor products in quantum state manipulation.

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QITStudent
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Hey guys, I need some help understanding some very basic quantum manipulation techniques.

If I am working out the bra version of a tensor product on some ket vectors, do the two ket vectors change from |v1>|v2> to <v1|<v2|, or to <v2|<v1|.

Or, to put in context, if I measure an operator with an eigenvector AxB on state |v1>|v2> , is the probability that I get the eigenvalue associated with this eigevector

<v1|<v2|AxB|v1>|v2> or <v2|<v1|AxB|v1>|v2>?​

Many thanks for looking through this with me
 
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If |v_1\rangle|v_2\rangle is a state is the tensor product \mathcal{H}_1 \otimes \mathcal{H}_2 of two Hilbert spaces, then the corresponding bra is in the tensor product of the dual spaces of the Hilbert spaces, namely \mathcal{H}_1^* \otimes \mathcal{H}_2^*. Thus, it is \langle v_1|\langle v_2|.
Btw, a more precise notation is |v_1\rangle\otimes|v_2\rangle\equiv|v_1\rangle|v_2\rangle and \langle v_1|\otimes\langle v_2|\equiv\langle v_1|\langle v_2|
 

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