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I am trying to solve the question below:

enter image description here

While solving the post measurement state, I understand we can take the 1st and last qubit common using tensor product if they are the same(1st part of the handwritten notes)

Is this true/correct to write for the qubits in the middle as well? That is, is the the correct representation of the system using tensor product (2nd part of the handwritten notes)?

My solution:

$$\lvert\psi\rangle = \frac{\sqrt 2 + i}{\sqrt{20}}\lvert 000\rangle + \frac{1}{\sqrt 2}\lvert 001\rangle + \frac{1}{\sqrt{10}}\lvert 011\rangle + \frac{i}{2}\lvert 111\rangle$$

  1. After measuring 1st qubit as $\lvert 0\rangle$, the state of the system is: $$\begin{align} \frac{\tfrac{\sqrt 2 + i}{\sqrt{20}}\lvert 000\rangle + \tfrac{1}{\sqrt 2}\lvert 001\rangle + \tfrac{1}{\sqrt{10}}\lvert 011\rangle}{\sqrt{|\tfrac{\sqrt 2 + i}{\sqrt{20}}|^2 + |\tfrac{1}{\sqrt 2}|^2 + |\tfrac{1}{\sqrt{10}}|^2}} &= \tfrac{\sqrt{2}+i}{\sqrt{15}}\lvert 000\rangle + \sqrt{\tfrac{2}{3}}\lvert 001\rangle +\sqrt{\tfrac{2}{15}}\lvert 011\rangle \\ &= \lvert 0\rangle \otimes \bigg(\tfrac{\sqrt{2}+i}{\sqrt{15}}\lvert 00\rangle + \sqrt{\tfrac{2}{3}}\lvert 01\rangle +\sqrt{\tfrac{2}{15}}\lvert 11\rangle\bigg)\end{align}$$

  2. After measuring the second qubit as $\lvert 1\rangle$, the state of the system is:

$$\begin{align} \frac{\tfrac{1}{\sqrt{10}}\lvert 011\rangle + \tfrac{i}{2}\lvert 111\rangle}{\sqrt{|\tfrac{1}{\sqrt{10}}|^2 + |\tfrac{i}{2}|^2}} &= \tfrac{1}{\sqrt{6}}\lvert 011\rangle + \tfrac{i\sqrt{5}}{2\sqrt{3}}\lvert 111\rangle \\ &=^? \bigg(\tfrac{1}{\sqrt{6}}\lvert 0\rangle + \tfrac{i\sqrt{5}}{2\sqrt{3}}\lvert 1\rangle\bigg)\otimes \lvert 1\rangle \otimes \bigg(\tfrac{1}{\sqrt 6}\lvert 1\rangle + \tfrac{i\sqrt{5}}{3}\lvert 1\rangle\bigg) \end{align}$$

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    $\begingroup$ Please use MathJax instead of posting pictures of large blocks of text... $\endgroup$
    – hft
    Mar 13, 2023 at 5:03
  • $\begingroup$ math.meta.stackexchange.com/questions/5020/… $\endgroup$
    – hft
    Mar 13, 2023 at 5:04
  • $\begingroup$ I am little new here, will keep this in mind. Thanks! $\endgroup$
    – Saniaaa
    Mar 13, 2023 at 5:08
  • $\begingroup$ @Saniaaa You should try to edit your post so it does not include images. $\endgroup$
    – Rammus
    Mar 13, 2023 at 7:13

1 Answer 1

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tl;dr: The approach is correct but OP's calculation features multiple errors.

Just to be precise about what I'll be doing: In the POVM formulation a quantum measurement is described by a collection $\{M_x\}_x$ of measurement operators (positive semi-definite operators, normalized to $\sum_x M_x={\bf1}$) indexed by the different measurement outcomes $x$. Now there are two versions of a post measurement state (cf., e.g., Chapter 3.3 in "Principles of Quantum Communication Theory: A Modern Approach" by Khatri & Wilde):

  • If the recorded outcome of the measurement was $x$, then the original state $\rho$ turned into the now post-measurement state $\frac{M_x\rho M_x^\dagger}{{\rm tr}(M_x\rho M_x^\dagger)}$. In case of a pure state $\psi$ this becomes $\frac{M_x|\psi\rangle}{\sqrt{\langle \psi|M_x^\dagger M_x|\psi\rangle}}=\frac{M_x|\psi\rangle}{\|M_x|\psi\rangle\|}$ (which is readily verified by substituting $\rho\to|\psi\rangle\langle\psi|$ in the first expression)
  • If the measurement outcome is not available (i.e. the experimenter does not have access to the measurement outcome), then one can still compute the expected density operator which is given by $\rho_M=\sum_x M_x\rho M_x^\dagger$. For the expected state one has to go to density operators because it can happen that the full system is in a pure state but the expected state is not pure anymore, but rather mixed.

While the first bullet point is probably what the exercise intends you to do I will showcase both procedures for completeness' sake. For measurements in the computational basis—assuming just one qubit for the moment—these measurement operators are given by $M_0=|0\rangle\langle 0|$ and $M_1=|1\rangle\langle 1|$, see, e.g., Chapter 2.2.3 in Nielsen & Chuang. Thus when measuring only the second of three subsystems this yields the measurement operators $M_0={\bf1}_A\otimes|0\rangle\langle 0|\otimes{\bf1}_C$ and $M_1={\bf1}_A\otimes|1\rangle\langle 1|\otimes{\bf1}_C$. This means:

  1. If only the second qubit is measured with outcome $1$, then the post-measurement state is \begin{align*} M_1|\psi\rangle&=\frac{({\bf1}_A\otimes|1\rangle\langle 1|\otimes{\bf1}_C)|\psi\rangle}{\|({\bf1}_A\otimes|1\rangle\langle 1|\otimes{\bf1}_C)|\psi\rangle\|}\\ &=\frac{\frac1{\sqrt{10}}|011\rangle+\frac i2|111\rangle}{\sqrt{\frac1{10}+\frac 14}}\\ &=\sqrt{\frac{20}7}\Big(\frac1{\sqrt{10}}|011\rangle+\frac i2|111\rangle\Big)\\ &=\sqrt{\frac27}|011\rangle+i\sqrt{\frac57}|111\rangle\,. \end{align*} As a sanity check this state is normalized because $$ \Big\|\sqrt{\frac27}|011\rangle+i\sqrt{\frac57}|111\rangle\Big\|=\sqrt{\frac27+\frac57}=1\,. $$ Interestingly this post-measurement state is indeed a product state as it factorizes as the second and third qubit are guaranteed to be in state $|1\rangle$: $$ \boxed{M_1|\psi\rangle=\Big(\sqrt{\frac27}|0\rangle+i\sqrt{\frac57}|1\rangle\Big)\otimes|1\rangle\otimes |1\rangle} $$
  2. For the expected state we compute \begin{align*} \rho_M&=M_0|\psi\rangle\langle\psi|M_0^\dagger+M_1|\psi\rangle\langle\psi|M_1^\dagger\\ &=\Big( \frac{\sqrt 2 + i}{\sqrt{20}}\lvert 000\rangle + \frac{1}{\sqrt 2}\lvert 001\rangle\Big)\Big( \frac{\sqrt 2 + i}{\sqrt{20}}\langle 000|+ \frac{1}{\sqrt 2}\langle001|\Big)\\ &\qquad+\Big( \frac{1}{\sqrt{10}}\lvert 011\rangle + \frac{i}{2}\lvert 111\rangle\Big) \Big( \frac{1}{\sqrt{10}}\langle011|+ \frac{i}{2}\langle111|\Big)\\ &=\Big( |0\rangle\otimes|0\rangle\otimes\Big(\frac{\sqrt 2 + i}{\sqrt{20}}\lvert 0\rangle + \frac{1}{\sqrt 2}\lvert 1\rangle\Big)\Big)\Big( \langle0|\otimes\langle0|\otimes\Big(\frac{\sqrt 2 + i}{\sqrt{20}}\langle0|+ \frac{1}{\sqrt 2}\langle1|\Big)\Big)\\ &\qquad+\Big( \Big(\frac{1}{\sqrt{10}}\lvert 0\rangle + \frac{i}{2}\lvert 1\rangle\Big) \otimes|1\rangle\otimes|1\rangle \Big) \Big( \Big(\frac{1}{\sqrt{10}}\langle0|+ \frac{i}{2}\langle1|\Big) \otimes\langle1|\otimes\langle1|\Big)\\ &= |0\rangle\langle 0|\otimes|0\rangle\langle 0|\otimes\Big( \frac{\sqrt 2 + i}{\sqrt{20}}\lvert 0\rangle + \frac{1}{\sqrt 2}\lvert 1\rangle \Big)\Big( \frac{\sqrt 2 + i}{\sqrt{20}}\langle0|+ \frac{1}{\sqrt 2}\langle1| \Big) \\ &\qquad+\Big( \frac{1}{\sqrt{10}}\lvert 0\rangle + \frac{i}{2}\lvert 1\rangle \Big) \Big( \frac{1}{\sqrt{10}}\langle0|+ \frac{i}{2}\langle1|\Big) \otimes|1\rangle\langle1|\otimes|1\rangle\langle1| \end{align*}
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    $\begingroup$ I edited the title of the question to try and make it more descriptive (and retrievable in the future). Let me know if you think it's not fully in line with your answer! $\endgroup$
    – glS
    May 22 at 14:13

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