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I have a doubt about how two equivalent ways of calculating the inner product between two states seem to not be actually equivalent, as they should. In particular, I'm interested in the case where the states belong to a two-dimensional state space and can be associated with a corresponding Bloch sphere.
To start with, let the two states be designated by $\left |\psi_{1}\right\rangle$ and $\left |\psi_{2}\right\rangle$. Now, in terms of the general angles $\theta$ and $\phi$ of the Bloch sphere, these two states can be reformulated in terms of a given basis $\{\left |0\right\rangle, \left |1\right\rangle\}$ as: $$\left |\psi_{1}\right\rangle=\cos\frac{\theta_{1}}{2}\left |0\right\rangle+\sin\frac{\theta_{1}}{2}e^{i\phi_{1}}\left |1\right\rangle, \left |\psi_{2}\right\rangle=\cos\frac{\theta_{2}}{2}\left |0\right\rangle+\sin\frac{\theta_{2}}{2}e^{i\phi_{2}}\left |1\right\rangle$$ and the modulus squared of their inner product can be written as: $$|\left\langle\psi_{2}|\psi_{1}\right\rangle|^{2}=|\cos\frac{\theta_{1}}{2}\cos\frac{\theta_{2}}{2}+\sin\frac{\theta_{1}}{2}\sin\frac{\theta_{2}}{2}e^{i(\phi_{1}-\phi_{2})}|^{2} \\ =\cos^{2}\frac{\theta_{1}}{2}\cos^{2}\frac{\theta_{2}}{2}+\sin^{2}\frac{\theta_{1}}{2}\sin^{2}\frac{\theta_{2}}{2}+2\cos\frac{\theta_{1}}{2}\cos\frac{\theta_{2}}{2}\sin\frac{\theta_{1}}{2}\sin\frac{\theta_{2}}{2}\cos(\phi_{2}-\phi_{1})$$ Now, an equivalent way of computing the same thing would be by determining the trace of the product of the density operators of the two states in the basis $\{\left |0\right\rangle, \left |1\right\rangle\}$, given that: $$|\langle\psi_2|\psi_{1}\rangle|^2=\langle\psi_{2}|\psi_{1}\rangle\langle\psi_{1}|\psi_{2}\rangle=\langle P_{\psi_{1}}\rangle_{\psi_{2}}=\operatorname{Tr}(\rho_{2}P_{\psi_{1}})=\operatorname{Tr}(\rho_{2}\rho_{1})$$ where $P_{\psi_{1}}$ is the projector on the first state, that is $P_{\psi_{1}}=|\psi_{1}\rangle\langle\psi_{1}|=\rho_{1}$. Using the result discussed in https://quantumcomputing.stackexchange.com/questions/6882/how-are-arbitrary-2-times-2-matrices-decomposed-in-the-pauli-basis, the two density matrices can be written in the basis $\{\left |0\right\rangle, \left |1\right\rangle\}$ in terms of the pauli matrices $\{\sigma_{i}\}$ as: $$\rho_{i}=\frac{1}{2}[\mathbb{I}+\vec{x}\cdot\vec{\sigma}],\ i=1,2$$ if computed this way the modulus squared of the inner product of the two states is: $$|\langle\psi_2|\psi_{1}\rangle|^2=\operatorname{Tr}(\rho_{2}\rho_{1})=\frac{1}{2}(1+\hat{e_{\psi_{1}}}\cdot\hat{e_{\psi_{2}}}) \\ =\frac{1}{2}[1+\sin(\theta_{1})\sin(\theta_{2})\cos(\phi_{2}-\phi_{1})+\cos(\theta_{1})\cos(\theta_{2}))]$$ where $\hat{e_{\psi_{i}}}$ is the Bloch vector of the state $|\psi_{i}\rangle$, that is, the unit vector whose end is the point on the surface of said sphere that represents that state: $$\hat{e_{\psi_{i}}}=(\sin(\theta_{i})\cos(\phi_{i}), \sin(\theta_{i})\sin(\phi_{i}), \cos(\theta_{i}))^{T}, i=1,2$$ Now, the result of this second way of computing the modulus squared of the inner product doesn't seem to coincide with the result obtained via the first method; the expressions are, to the best of my knowledge, not equivalent. How is this possible?

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    $\begingroup$ Have you tried using standard half-angle trig identities to show that your two formulae coincide (or not as the case may be). $\endgroup$
    – mike stone
    Commented May 8 at 13:45
  • $\begingroup$ Hi, thanks for answering. I have tried indeed, but I still don´t manage to get the same result as when I use the density matrices, which is apparently the right result (I´m referencing Mittelstaedt "The interpretation of Quantum Mechanics and the Measurement process" p.37, in which this result is exhibited via the density matrix formalismo but not the state formalism; this is essentially where my question comes from). If you get any ideas, let me know. Also, how would it be possible the two formulae dont coincide? Warm greetings. $\endgroup$ Commented May 8 at 13:55

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As you pointed out both approaches are equivalent meaning they have to lead to the same answer. As hinted at by Mike Stone in their comment the two expressions can be related via the following basic trigonometric identities:

  • $\sin\theta_j=2\sin\frac{\theta_j}2\cos\frac{\theta_j}2$
  • $(\sin\frac{\theta_j}2)^2=\frac{1-\cos\theta_j}2$
  • $(\cos\frac{\theta_j}2)^2=\frac{1+\cos\theta_j}2$

Based on your (correct) computation for $|\langle\psi_{2}|\psi_{1}\rangle|^2$ this yields \begin{align*} |\left\langle\psi_{2}|\psi_{1}\right\rangle|^{2}&=\Big(\cos\frac{\theta_{1}}{2}\Big)^2\Big(\cos\frac{\theta_{2}}{2}\Big)^2+\Big(\sin\frac{\theta_{1}}{2}\Big)^2\Big(\sin\frac{\theta_{2}}{2}\Big)^2\\ &\qquad\qquad\qquad+\frac12\Big(2\sin\frac{\theta_{1}}{2}\cos\frac{\theta_{1}}{2}\Big)\Big(2\cos\frac{\theta_{2}}{2}\sin\frac{\theta_{2}}{2}\Big)\cos(\phi_{2}-\phi_{1})\\ &= \frac{1+\cos\theta_1}2\frac{1+\cos\theta_2}2+\frac{1-\cos\theta_1}2\frac{1-\cos\theta_2}2+\frac12 \sin\theta_1\sin\theta_2\cos(\phi_2-\phi_1)\\ &=\frac{2+2\cos\theta_1\cos\theta_2}4+\frac12 \sin\theta_1\sin\theta_2\cos(\phi_2-\phi_1)\\ &=\frac12\big( 1+\cos\theta_1\cos\theta_2+\sin\theta_1\sin\theta_2\cos(\phi_2-\phi_1) \big)\,. \end{align*} But this is exactly what $\frac{1}{2}(1+\hat{e_{\psi_{1}}}\cdot\hat{e_{\psi_{2}}})$ yielded so we are done.

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  • $\begingroup$ Hi, Frederik Vom Ende. I'm very thankful for your response. In the end, I manage to get the same result both ways; I suppose I must have made some mistake that I dind't notice at the time of posting this. I appreciate you taking the time to adress my question anyways. Warm Greetings. $\endgroup$ Commented May 14 at 18:31

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