Inner product in a composite Hilbert space

Question

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 :

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

Answer 1

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

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

$(⟨1:𝑛′|⊗⟨2:𝑝′|)|(|1:𝑛⟩⊗|2:𝑝⟩)= ⟨1:𝑛′|1:𝑛⟩⊗⟨2:𝑝′|2:𝑝⟩ =⟨1:𝑛′|1:𝑛⟩⟨2:𝑝′|2:𝑝⟩$,

where in the last equality the tensor product transforms into scalar product since $⟨1:𝑛′|1:𝑛⟩$ and $⟨2:𝑝′|2:𝑝⟩$ 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

$⟨2:𝑝′|(|1:𝑛⟩⊗|2:𝑝⟩)=|1:𝑛⟩⊗⟨2:𝑝′|2:𝑝⟩=|1:𝑛⟩⟨2:𝑝′|2:𝑝⟩$.

This is exactly what you should obtain, namely, the second system is traced out by projecting it on $|2:𝑝⟩$, while the first system remains untouched, which is given by the vector $|1:𝑛⟩$.