# Inner product in a composite Hilbert space

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 :

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

• This is the usual abuse of notation regarding the partial trace, I think. Have a look here, for example. Dec 24, 2022 at 14:22
• ah ok , so $\langle 2 : + | + - \rangle$ is really $( I_{d1} \otimes \langle 2 : + | ) |+ - \rangle$. So $( I_{d1}+ \langle +| ) |+ - \rangle = |1 : +\rangle ( \langle 2:+ | 2: - \rangle )$ Dec 24, 2022 at 14:35
• Sorry, it is hard for me to read this notation (especially in a comment). Dec 24, 2022 at 14:40
• $$\left(\mathbb I_A \otimes \langle \psi| \right) (|\varphi\rangle \otimes|\phi\rangle) = |\varphi\rangle \langle \psi|\phi \rangle \quad$$ is what I meant to say Dec 24, 2022 at 14:42
• 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). Dec 24, 2022 at 14:46

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 $$|1:𝑛⟩$$.

• Consider to use MathJax. Dec 25, 2022 at 10:58
• Noted, thanks. @Tobias Fünke Dec 25, 2022 at 19:06