Why isn't the circuit performing a measurement in the Bell basis? - Quantum Computing Stack Exchange most recent 30 from quantumcomputing.stackexchange.com 2022-01-25T22:31:46Z https://quantumcomputing.stackexchange.com/feeds/question/5605 https://creativecommons.org/licenses/by-sa/4.0/rdf https://quantumcomputing.stackexchange.com/q/5605 4 Why isn't the circuit performing a measurement in the Bell basis? bilanush https://quantumcomputing.stackexchange.com/users/5742 2019-03-03T10:11:27Z 2019-03-04T07:58:40Z <p>Nielsen and Chuang (on page 188 exercise 4.33) says that the circuit including CNOT and Hadamard is performing a measurement in the Bell basis. But I can't see how. </p> <p><a href="https://i.stack.imgur.com/nHiYP.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/nHiYP.png" alt="enter image description here"></a></p> <p>The matrix representing the circuit functionality is:</p> <p><span class="math-container">$$\begin{bmatrix} 1 &amp; 0 &amp; 0 &amp; 1 \\ 0 &amp; 1 &amp; 1 &amp; 0 \\ 1 &amp; 0 &amp; 0 &amp; -1 \\ 0 &amp; 1 &amp;-1 &amp; 0 \end{bmatrix}$$</span></p> <p>Now, what I have found out is, that instead of measuring in the Bell state, it is actually simply converting <span class="math-container">$|00\rangle$</span> to <span class="math-container">$\beta_{00}$</span>, <span class="math-container">$|01\rangle$</span> to <span class="math-container">$\beta_{01}$</span> and vice versa and so on. So it's actually transforming from one basis to another. But how is it a measurement in the Bell state?</p> <p>Suppose you take the vector <span class="math-container">$(1,0,0,0)$</span> in the Bell state it's written <span class="math-container">$(1,0,1,0)$</span> then when measuring in Bell state I would expect you get either <span class="math-container">$(1,0,0,0)$</span> which is <span class="math-container">$|00\rangle + |11\rangle$</span> or <span class="math-container">$(0,0,1,0)$</span> which is <span class="math-container">$|00\rangle-|11\rangle$</span>. But why isn't it doing the measurement but just converting from basis to basis and yet Wikipedia says it's a measurement? </p> https://quantumcomputing.stackexchange.com/questions/5605/-/5606#5606 6 Answer by Sanchayan Dutta for Why isn't the circuit performing a measurement in the Bell basis? Sanchayan Dutta https://quantumcomputing.stackexchange.com/users/26 2019-03-03T11:01:35Z 2019-03-03T13:53:27Z <ol> <li>You're correct that unitary gates like the Hadamard and CNOT <em>do not</em> perform any measurement. In fact, measurement isn't a unitary operation at all! There are separate measurements gates to measure the individual qubit states. The basis which the measurement gates will use is implementation dependent. However, in your specific circuit (N&amp;C p. 188, Ex. 4.33) they're using the <span class="math-container">$\{|0\rangle, |1\rangle\}$</span> basis. Notice the <a href="https://i.stack.imgur.com/6bAWo.png" rel="noreferrer">meter type gates</a> in your circuit. Those are the measurement gates. That circuit is essentially the one used to distinguish between the four orthogonal Bell states. They're mapped to <span class="math-container">$|00\rangle$</span>, <span class="math-container">$|01\rangle$</span>, <span class="math-container">$|10\rangle$</span> and <span class="math-container">$|11\rangle$</span> respectively. The mappings are (notations explained <a href="https://en.wikipedia.org/wiki/Bell_state#Bell_basis" rel="noreferrer">here</a>):</li> </ol> <p><span class="math-container">$$|\Phi^+\rangle \to |00\rangle$$</span> <span class="math-container">$$|\Phi^-\rangle \to |10\rangle$$</span> <span class="math-container">$$|\Psi^+\rangle \to |01\rangle$$</span> <span class="math-container">$$|\Psi^-\rangle \to |11\rangle$$</span></p> <ol start="2"> <li>As for how we can distinguish the four Bell states using two measurement gates, read <a href="https://physics.stackexchange.com/users/911/craig-gidney">Craig Gidney's</a> answer to <a href="https://physics.stackexchange.com/questions/285878/distinguishing-between-orthogonal-bell-states">Distinguishing between orthogonal bell states</a>.</li> </ol> <blockquote> <p>Typically, a Bell basis measurement is done <a href="http://algorithmicassertions.com/quirk#circuit=%7B%22cols%22%3A%5B%5B%22Counting2%22%5D%2C%5B%22H%22%5D%2C%5B%22%E2%80%A2%22%2C%22X%22%5D%2C%5B%22%E2%80%A6%22%2C%22%E2%80%A6%22%5D%2C%5B%22Amps2%22%5D%2C%5B%5D%2C%5B%22%E2%80%A2%22%2C%22X%22%5D%2C%5B%22H%22%5D%2C%5B%22Chance%22%2C%22Chance%22%5D%5D%7D" rel="noreferrer">with a CNOT and a Hadamard like so</a>:</p> <p><a href="https://i.stack.imgur.com/aZhyS.png" rel="noreferrer"><img src="https://i.stack.imgur.com/aZhyS.png" alt="Bell basis measurement"></a></p> <p>It's exactly the reverse of making a Bell pair.</p> <p>You can also <a href="http://algorithmicassertions.com/quirk#circuit=%7B%22cols%22%3A%5B%5B%22Counting2%22%5D%2C%5B%22H%22%5D%2C%5B%22%E2%80%A2%22%2C%22X%22%5D%2C%5B%22%E2%80%A6%22%2C%22%E2%80%A6%22%5D%2C%5B%22Amps2%22%5D%2C%5B%5D%2C%5B%22%E2%80%A2%22%2C1%2C%22X%22%5D%2C%5B1%2C%22%E2%80%A2%22%2C%22X%22%5D%2C%5B%22%E2%8A%95%22%2C1%2C1%2C%22X%22%5D%2C%5B1%2C%22%E2%8A%95%22%2C1%2C%22X%22%5D%2C%5B1%2C1%2C%22Measure%22%2C%22Measure%22%5D%2C%5B%22Amps2%22%2C1%2C%22Chance%22%2C%22Chance%22%5D%5D%7D" rel="noreferrer">simply do parity measurements along the X and Z axes</a>:</p> <p><a href="https://i.stack.imgur.com/xMjyN.png" rel="noreferrer"><img src="https://i.stack.imgur.com/xMjyN.png" alt="Bell basis parity measurement"></a></p> </blockquote> <hr> <p><strong>Further Reading:</strong> </p> <ol> <li><p><a href="https://quantumcomputing.stackexchange.com/questions/136/if-all-quantum-gates-must-be-unitary-what-about-measurement/143">If all quantum gates must be unitary, what about measurement?</a></p></li> <li><p><a href="https://quantumcomputing.stackexchange.com/questions/1870/what-does-measurement-in-a-certain-basis-mean">What does “measurement in a certain basis” mean?</a></p></li> <li><p><a href="https://physics.stackexchange.com/questions/261664/how-to-measure-relative-phases-of-quantum-states">How to measure relative phases of quantum states?</a></p></li> </ol> https://quantumcomputing.stackexchange.com/questions/5605/-/5613#5613 1 Answer by DaftWullie for Why isn't the circuit performing a measurement in the Bell basis? DaftWullie https://quantumcomputing.stackexchange.com/users/1837 2019-03-04T07:58:40Z 2019-03-04T07:58:40Z <p>Just to supplement the given answer, if you want to know what the effective measurement is, you start from the measurement operators and work backwards. So, the two measurements mean that you're projecting on the states <span class="math-container">$|00\rangle,|01\rangle,|10\rangle,|11\rangle$</span>. How do these transform under the preceding gates? Immediately before the measurement, we have a Hadamard on the first quibt. Incorporating that into the measurement basis means measuring in <span class="math-container">$|+0\rangle,|+1\rangle,|-0\rangle,|-1\rangle$</span>. Finally, we incorporate the controlled-not, and we see that the overall thing corresponds to measurement in the Bell basis.</p>