Can everything in QM be described with degrees instead of matrices and vectors? - Quantum Computing Stack Exchange most recent 30 from quantumcomputing.stackexchange.com 2022-01-26T00:09:57Z https://quantumcomputing.stackexchange.com/feeds/question/8390 https://creativecommons.org/licenses/by-sa/4.0/rdf https://quantumcomputing.stackexchange.com/q/8390 2 Can everything in QM be described with degrees instead of matrices and vectors? guest https://quantumcomputing.stackexchange.com/users/8491 2019-10-03T05:06:03Z 2021-01-03T11:09:16Z <p>I found <a href="https://www.quantum-inspire.com/kbase/hadamard/" rel="nofollow noreferrer">this</a> explanation.</p> <p>"The Hadamard gate can also be expressed as a 90º rotation around the Y-axis, followed by a 180º rotation around the X-axis. So <span class="math-container">$H=XY^{1/2}H = X Y^{1/2}H=XY^{1/2}$</span>."</p> <p>Can everything in QM be explained with degrees instead of matrices and vectors?</p> https://quantumcomputing.stackexchange.com/questions/8390/-/8396#8396 1 Answer by John for Can everything in QM be described with degrees instead of matrices and vectors? John https://quantumcomputing.stackexchange.com/users/5998 2019-10-03T10:05:49Z 2019-10-03T10:05:49Z <p>The state of single qubit can be described as a point on the <a href="https://en.wikipedia.org/wiki/Bloch_sphere" rel="nofollow noreferrer">Bloch sphere</a>. All the allowed transformations of a single qubit can then be described as rotations on the Bloch sphere. Unfortunately, bigger quantum systems can no longer be described as fitting on a sphere like geometry. As a result, this idea of considering transformations as rotations does not hold for all quantum systems.</p> https://quantumcomputing.stackexchange.com/questions/8390/-/8400#8400 0 Answer by Faisal Debouni for Can everything in QM be described with degrees instead of matrices and vectors? Faisal Debouni https://quantumcomputing.stackexchange.com/users/8647 2019-10-03T13:06:47Z 2019-10-09T06:52:27Z <p>Short answer: Yes, except for measurement.</p> <p>There two postulates in play here:<br> 1- the evolution of a closed QM system can always be described by a unitary matrix.<br> 2- measurement operators (observable operator) are always hermitian</p> <p>Hermitian operators can't (always) be described as a rotation in any space.</p> <p>Unitary operators can always be considered as rotation around some axis since they always preserve the inner products. However the dimensions of the space where the rotation happens increases as the dimension of the system increase: </p> <ul> <li>operation on 1 qubit is a rotation on the 2D surface of a 3D sphere (Bloch sphere). </li> <li>operation on m qubits is a rotation on the <span class="math-container">$2^{m+1}$</span> − 2 dimensions surface of some <span class="math-container">$2^{m+1}$</span> − 1 sphere. </li> </ul> <p>Which makes this visualization (degrees of rotation) not very helpful when applied to a multi-qubit system.</p> https://quantumcomputing.stackexchange.com/questions/8390/-/8455#8455 2 Answer by glS for Can everything in QM be described with degrees instead of matrices and vectors? glS https://quantumcomputing.stackexchange.com/users/55 2019-10-10T16:43:16Z 2019-10-10T16:43:16Z <p>In principle, yes, you can always do it. The Bloch representation can be generalised to arbitrary dimensions, and you can always parametrise states in it by their "angle coordinates".</p> <p>For example, you can write an arbitrary 3-modes pure state as <span class="math-container">$$|\psi\rangle=\cos\alpha|0\rangle + e^{i\theta}\sin\alpha\cos\beta|1\rangle+e^{i\phi}\sin\alpha\sin\beta|2\rangle,$$</span> for <span class="math-container">$\alpha,\beta,\theta,\phi\in\mathbb R$</span>.</p> <p>It should be noted, however, that things get much with spaces of dimension larger than <span class="math-container">$2$</span>. For example, it's harder to interpret arbitrary unitary gates as <em>rotations</em> in this larger space. By this, I mean that even if it is always true that for any given unitary <span class="math-container">$U$</span> there is some (and in general an infinity of) Hermitians <span class="math-container">$H$</span> such that <span class="math-container">$e^{-iHt}=U$</span> at some time <span class="math-container">$t$</span>, whether these should be considered "rotations" is arguable.</p> <p>On the one hand, if you represent <span class="math-container">$H$</span> as a point in the Bloch representation (you can always do this because the Bloch representation associates a point to any Hermitian matrix, even though this point will in general fall outside of the region representing the set of physical states), then you can think of <span class="math-container">$H$</span> as the "axis" of the rotation, as the direction associated with this point will be fixed by the rotation.</p> <p>On the other hand, in general <span class="math-container">$t\mapsto e^{-iHt}$</span> is not a rotation, in the sense that it doesn't "loop back" as rotations do in the Bloch sphere. By this I mean that in general there is no <span class="math-container">$t&gt;0$</span> such that <span class="math-container">$e^{-iHt}=I$</span>, which is what you would expect from a rotation. An easy example of this is: <span class="math-container">$$H\equiv \begin{pmatrix}\alpha&amp;0&amp;0\\0&amp;\beta&amp;0\\0&amp;0&amp;\gamma\end{pmatrix},\quad\alpha,\beta,\gamma\in\mathbb R.$$</span> Then, <span class="math-container">$e^{-iHt}=\mathrm{diag}(e^{-i\alpha t},e^{-i\beta t},e^{-i\gamma t})$</span>, and <span class="math-container">$e^{-iH t}=I$</span> if and only if <span class="math-container">$e^{-i\alpha t}=e^{-i\beta t}=e^{-i\gamma t}=1$</span>, which aren't simultaneously satisfiable for incommensurable <span class="math-container">$\alpha,\beta,\gamma$</span>.</p>