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Take the two Hilbert spaces $ H_1 = H_2 = C^2$

The basis of $H_1$ is : $ \{ | 1 : + \rangle , |1 : - \rangle \} $ and for $H_2$ : $ \{ | 2 : + \rangle , |2 : - \rangle \} $

Forming the composite Hilbert space: $$ H = H_1 \otimes H_2 $$

We get the base of $H$ : $ \{ | 1 : + \rangle \otimes | 2 : + \rangle , | 1 : + \rangle \otimes | 2 : - \rangle , | 1 : - \rangle \otimes | 2 : - \rangle, | 1 : - \rangle \otimes | 2 : + \rangle \} $. Written more simply as $$ \{| + + \rangle , | + - \rangle , | - - \rangle , |- + \rangle \} $$

Let the composite system be in a ket $$ | \Psi \rangle = \dfrac{ |+ - \rangle - | - + \rangle}{\sqrt{2}} $$

I wish to calculate the partial trace of the density operator with respect to $H_2$

$$ \rho_1 = tr_2 \rho = tr_2 | \Psi \rangle \langle \Psi| = \langle 2 : + | \Psi \rangle \langle \Psi | 2 : + \rangle + \langle 2 : - | \Psi \rangle \langle \Psi | 2 : - \rangle $$

I cannot get past this step , since I don't know what: $$\langle 2: + | + - \rangle ?$$

Is it just $ \langle 2: + | 2: - \rangle = 0 $ ?

I know that for a vector space that is the tensor product of two other vector spaces the scalar product is :

$$ (\langle 1:n' |\otimes \langle 2: p' |)| (|1 : n \rangle \otimes |2:p \rangle) = \ \langle 1:n'| 1 : n \rangle \times \langle 2: p' | 2:p \rangle $$

What is then :

$$ \langle 2: p' | (|1 : n \rangle \otimes |2:p \rangle) ? $$

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  • $\begingroup$ This is the usual abuse of notation regarding the partial trace, I think. Have a look here, for example. $\endgroup$ Dec 24, 2022 at 14:22
  • $\begingroup$ ah ok , so $ \langle 2 : + | + - \rangle$ is really $( I_{d1} \otimes \langle 2 : + | ) |+ - \rangle $. So $( I_{d1}+ \langle +| ) |+ - \rangle = |1 : +\rangle ( \langle 2:+ | 2: - \rangle ) $ $\endgroup$
    – lohey
    Dec 24, 2022 at 14:35
  • $\begingroup$ Sorry, it is hard for me to read this notation (especially in a comment). $\endgroup$ Dec 24, 2022 at 14:40
  • $\begingroup$ $$ \left(\mathbb I_A \otimes \langle \psi| \right) (|\varphi\rangle \otimes|\phi\rangle) = |\varphi\rangle \langle \psi|\phi \rangle \quad $$ is what I meant to say $\endgroup$
    – lohey
    Dec 24, 2022 at 14:42
  • $\begingroup$ Yes, indeed. As I've explained in the linked answer, the notation you encountered (I suppose), is a common abuse of notation (cf. eq. 4 there). $\endgroup$ Dec 24, 2022 at 14:46

1 Answer 1

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Using your notation, you already pointed out that ${|2:+⟩,|2:βˆ’βŸ©}$ form a basis in $𝐻_2$, so $⟨2:+|2:βˆ’βŸ©$ is definitely zero.

Second, you also observed the rule by which the inner product and tensor product interchange, i.e.

$(⟨1:𝑛′|βŠ—βŸ¨2:𝑝′|)|(|1:π‘›βŸ©βŠ—|2:π‘βŸ©)= ⟨1:𝑛′|1:π‘›βŸ©βŠ—βŸ¨2:𝑝′|2:π‘βŸ© =⟨1:𝑛′|1:π‘›βŸ©βŸ¨2:𝑝′|2:π‘βŸ©$,

where in the last equality the tensor product transforms into scalar product since $⟨1:𝑛′|1:π‘›βŸ©$ and $⟨2:𝑝′|2:π‘βŸ©$ are just numbers.

Exactly the same rule applies to your last equation, the only difference is that there you have an inner product only between vectors of $𝐻_2$, while the vector of $𝐻_1$ remains unchanged

$⟨2:𝑝′|(|1:π‘›βŸ©βŠ—|2:π‘βŸ©)=|1:π‘›βŸ©βŠ—βŸ¨2:𝑝′|2:π‘βŸ©=|1:π‘›βŸ©βŸ¨2:𝑝′|2:π‘βŸ©$.

This is exactly what you should obtain, namely, the second system is traced out by projecting it on $|2:π‘βŸ©$, while the first system remains untouched, which is given by the vector $|1:π‘›βŸ©$.

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  • $\begingroup$ Consider to use MathJax. $\endgroup$ Dec 25, 2022 at 10:58
  • $\begingroup$ Noted, thanks. @Tobias Fünke $\endgroup$
    – tamih100
    Dec 25, 2022 at 19:06

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