This paper states:

Suppose a two qubit system is in the state $|\psi\rangle=a|00\rangle+b|11\rangle$, and consider the expectation value of any observable $A \otimes I$ that is nontrivial only on the first factor: $\langle\psi|A \otimes I| \psi\rangle=|a|^{2}\langle 00|A \otimes I| 00\rangle+|b|^{2}\langle 11|A \otimes I| 11\rangle+a^{*} b\langle 00|A \otimes I| 11\rangle+b^{*} a\langle 11|A \otimes I| 00\rangle$ $$ =|a|^{2}\langle 0|A| 0\rangle+|b|^{2}\langle 1|A| 1\rangle $$

What is the significance $I$ in "$|A \otimes I|$"? I understand it is related to "on the first factor" but not quite sure how.


1 Answer 1


$I$ is the identity operator applied to the second subsystem, making it a trivial operation, ie one that doesn't affect it. Hence the expectation value of such an operator is only related to the expectation of the non-trivial operator, $A$ in this case, on it's subspace.

$$\langle \Psi|A \otimes I|\Psi\rangle=Tr(A\otimes I |\Psi\rangle\langle\Psi|)=Tr(ATr_{B}(|\Psi\rangle\langle\Psi|))=Tr(A\rho)$$ where $\rho=|a|^{2}|0\rangle\langle0|+|b|^{2}|1\rangle\langle1|$

  • $\begingroup$ Thank you so much. By "Hence the expectation value of such an operator" you meant the Identity $I$? . Also why is the expectation of A non-trivial? Only because of the tensor product with $I$? Also, first factor means the first qubit of A? $\endgroup$
    – mish
    Feb 17, 2022 at 15:16
  • $\begingroup$ By "such an operator" I mean $A \otimes I$. The expectation of $A$ is non-trivial because it can vary depedning on $A$, while $I$ will always be fixed. Not the first qubit of $A$. It's an operator, not a system. The first qubit of $|\Psi\rangle$. $\endgroup$ Feb 19, 2022 at 15:26

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