Quantum systems with real structure - Physics Stack Exchange most recent 30 from physics.stackexchange.com 2019-08-20T21:01:17Z https://physics.stackexchange.com/feeds/question/17589 http://www.creativecommons.org/licenses/by-sa/3.0/rdf https://physics.stackexchange.com/q/17589 7 Quantum systems with real structure Christoph https://physics.stackexchange.com/users/6389 2011-11-28T19:29:07Z 2011-11-29T18:13:05Z <p>A lot has been said as to why quantum mechanics needs complex numbers.</p> <p>However, all measurements produce real values. Expectation values are real, the observables form a real Lie-algebra (use $-\mathrm i[&#183;,&#183;]$ as Lie bracket instead of the raw commutator) and any state $\mathbf z$ is uniquely determined by the real-valued function $P_{\mathbf z}(&#183;) = |\langle \mathbf z,&#183;\rangle|^2$ on the space of states.</p> <p>The space of states itself, however, is a complex projective space, and even though phase factors are irrelevant when taking expectation values, they do matter when combining states - linear superposition $α\mathbf z + β\mathbf z&#39;$ depends on the ration $|α|/|β|$ as well as the phase difference $\arg(α) - \arg(β)$.</p> <p>But there is at least one important quantum system which can be modeled as a real system: the qubit, ie the space of spin states of a single electron.</p> <p>In addition to the standard representation as $\mathbf z=(z_1,z_2)\in\mathbb C^2,\langle \mathbf z,\mathbf z\rangle =1$, states can also be represented as $\vec z\in\mathbb R^3,\vec z^2=1$ via $$\vec z = \begin{pmatrix} \langle\mathbf z,\pmb σ_1\mathbf z\rangle\\ \langle\mathbf z,\pmb σ_2\mathbf z\rangle\\ \langle\mathbf z,\pmb σ_3\mathbf z\rangle \end{pmatrix} = \begin{pmatrix} 2\Re(\bar z_1z_2)\\ 2\Im(\bar z_1z_2)\\ |z_1|^2 - |z_2|^2 \end{pmatrix}$$ where $\pmb σ_1, \pmb σ_2, \pmb σ_3$ are the Pauli-matrices.</p> <p>This is the well-known Hopf fibration.</p> <p>What makes this useful is the fact that $P_z$ takes a particular simple form: $$P_{\mathbf z}(\mathbf z&#39;) = \frac 12 (1 + \vec z&#183;\vec z&#39;) = \frac 12 (1 + \cos\measuredangle(\vec z,\vec z&#39;))$$</p> <p>This also means that states $\mathbf z,\mathbf z&#39;$ are orthogonal in $\mathbb C^2$ exactly when $\measuredangle(\vec z,\vec z&#39;)=180&#176;$ in $\mathbb R^3$, ie $\vec z&#39;=-\vec z$</p> <p>The expection value of any observable $\mathbf A$ can be expanded in terms of $P_{\mathbf a_i}$ after a choice of orthonormal eigenvectors $\{\mathbf a_i\}$, ie $$\langle\mathbf z,\mathbf{Az}\rangle = \sum_i α_iP_{\mathbf a_i}(\mathbf z)$$ where $\{α_i\}$ denote the eigenvalues.</p> <p>Substituting our definition of $P_{\mathbf a_i}$ and using the fact that $\mathbf a_1=-\mathbf a_2$, this is simply $$\langle\mathbf z,\mathbf{Az}\rangle = \frac 12(α_1 + α_2) + \frac 12(α_1-α_2)\,\vec a_1&#183;\vec z$$</p> <p>Restricting ourselves to $\mathfrak{su}(2)$, ie the traceless matrices with $α_2=-α_1$, yields $$\langle\mathbf z,\mathbf{Az}\rangle = \vec A&#183;\vec z$$ where $\vec A=α_1\,\vec a_1$ is the representation of our observable.</p> <p>We arrived at a quantum system which can be described with nothing more than school math.</p> <p>After this lengthy exposition, finally my question: Are there any other quantum system with a similar real structure? If not, is there a particular reason why?</p> <h3>Update</h3> <p>To get a better understanding of the problem, I read up on complex manifolds, and it seems that the qubit is indeed special.</p> <p>In particular, the projective spaces $P_n\mathbb C$ <em>are</em> complex manifolds, but spheres $S^k$ do not even permit almost-complex structures for $k\not= 2,6$ (Borel, Serre 1951) and are out as state spaces in the general case.</p> <p>The finite-dimensional complex projective spaces can be realized as different homogeneous spaces. In particular, we have the obvious $$P_n\mathbb C \cong \mathbb C^{n+1}\setminus\{0\}\:\big/\:\mathbb C^*$$ the not-so-obvious $$P_n\mathbb C \cong U(n+1)\:\big/\:U(n)\times U(1)$$ and the one which is probably the most clear-cut $$P_n\mathbb C \cong S^{2n+1}\:\big/\:S^1$$ but does not convey that we are dealing with complex spaces. This quotient also only ever results in a sphere for $n=1$, ie our qubit case.</p> <p>The Wikipedia page on <a href="http://en.wikipedia.org/wiki/Hopf_fibration" rel="nofollow">Hopf fibration</a> links to <a href="http://arxiv.org/abs/quant-ph/0108137" rel="nofollow">this paper</a>, where the fibration $$S^7\xrightarrow{S^3}S^4$$ is used to model the state space $P_3\mathbb C$ of two qubits. Even though there <em>might</em> be some insight into the structure of the state space in this fibration, it is less natural (and less useful, for that matter) than the single qubit Hopf fibration $$S^3\xrightarrow{S^1}S^2$$ where actually $S^2 \cong P_1\mathbb C$, whereas obviously $S^4\not\cong P_3\mathbb C$ as they do not even have the same dimensions when considered as real manifolds.</p> https://physics.stackexchange.com/questions/17589/-/17590#17590 10 Answer by Ron Maimon for Quantum systems with real structure Ron Maimon https://physics.stackexchange.com/users/4864 2011-11-28T19:43:22Z 2011-11-28T19:43:22Z <p>If you replace the complex number "i" with the real 2 by 2 matrix I = (0,-1;1,0), and assert that every observable commutes with I, and replace i with I in all formulas, you get the pure real formulation of quantum mechanics. This is no more mysterious than writing a complex number as two real numbers, and this is exactly what you did.</p> <p>The real question of "why complex numbers" is why nearly every observable commutes with I. The "I" matrix is the real thing in quantum mechanics, not the algebraic "i" (which is equivalent). The only thing that does not commute with "i" is the time reversal operator. You can define I to commute with the Hamiltonian, the question is why it also commutes with everything else.</p>