# Mathematics of Measurement then Partial Trace

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?

• presumably you should have $\Pi_1\otimes I$, rather than just $\Pi_1$, also in the numerator, in the second equation?
– glS
Commented Jan 6, 2022 at 10:02

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.
• 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? Commented Jan 5, 2022 at 19:12
• @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 Commented Jan 5, 2022 at 20:23