1
$\begingroup$

Say we have the following quantum state: $$ |\psi\rangle = \frac{1}{\sqrt{2}}(|00\rangle +|10\rangle)$$

To measure the first qubit and then further trace out the first qubit, my notes have the following equation for the post-measurement state:

$$ \rho = \frac{\text{Tr}_1(\Pi_1|\psi\rangle\langle\psi|)}{\text{Tr}(\Pi_1\otimes \mathbb{I} ​|\psi\rangle\langle\psi|)}$$

Firstly, if I wanted to measure the first qubit in the state $|0\rangle$, what would $\Pi_1$ be? Would it be the outer product $|0\rangle\langle0|$ because $|\psi\rangle$ is in outer product form?

Secondly, what would the mathematics look like to get the post-measurement state?

$\endgroup$
1
  • 1
    $\begingroup$ presumably you should have $\Pi_1\otimes I$, rather than just $\Pi_1$, also in the numerator, in the second equation? $\endgroup$
    – glS
    Jan 6 at 10:02

1 Answer 1

1
$\begingroup$

Yes, this is exactly what happens. Instead of saying "I want to measure this specific value," you specify a set of POVM elements asking "which of these values will I get?". When you are measuring the value of the first qubit alone to see if it is in state $|0\rangle$ or $|1\rangle$, the two POVM elements can be specified by $\Pi^{(0)}=|0\rangle\langle 0|\otimes \mathbb{I}$ and $\Pi^{(1)}=|1\rangle\langle 1|\otimes \mathbb{I}$. Then, if you measure the first qubit to be $|0\rangle$, meaning that your measurement device records some binary piece of information telling you that it has measured the first qubit to be $|0\rangle$, you follow the state-update rule provided with $\Pi^{(0)}$.

You should be able to do the rest from here! Just remember that the trace in the denominator can easily be manipulated into $$\mathrm{Tr}(\Pi^{(0)}|\psi\rangle\langle\psi|)=\langle\psi|\Pi^{(0)}|\psi\rangle=\langle\psi|(|0\rangle\langle 0|\otimes \mathbb{I})|\psi\rangle$$ but you cannot do the same with the partial trace in the numerator. For the partial trace, you need to remember the rule $$\mathrm{Tr}_1(X_{12})=(\langle 0|_1\otimes \mathbb{I}_2)X_{12}(| 0\rangle_1\otimes \mathbb{I}_2)+(\langle 1|_1\otimes \mathbb{I}_2)X_{12}(| 1\rangle_1\otimes \mathbb{I}_2).$$ In this last expression, I am explicitly using subscripts to indicate to which Hilbert space each item belongs - you can match all of the subscripts for each of the qubit states and the operators to do the calculations.

$\endgroup$
2
  • $\begingroup$ So for the numerator, would I write $\text{Tr}_1 (|0\rangle\langle0| |\psi\rangle\langle\psi|)$ or $\text{Tr}_1 (|0\rangle\langle0| \otimes \mathbb{I} |\psi\rangle\langle\psi|)$. Furthermore, what does $X_{12}$ represent in your last equation? $\endgroup$ Jan 5 at 19:12
  • 1
    $\begingroup$ @JamesEllis $X_{12}$ represents any operator that acts on both qubits (it could be $|\psi\rangle\langle\psi|$, it could be $(|0\rangle_1\langle\otimes \mathbb{I}_2)|\psi\rangle\langle\psi|$, etc. You can always write $|0\rangle_1\langle0|$ if you remember that this automatically implies you are doing nothing (ie, the identity operation) on the remaining qubits. When learning things I like to make everything explicit and put in the identity operators, but in general most people leave them out when the context explains things $\endgroup$ Jan 5 at 20:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.