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Say I have a state, $$| \Psi \rangle = \frac{1}{\sqrt 2} \left( | 0 \rangle + \exp( \text{i} \phi ) | 1 \rangle \right) = c_{0} | 0 \rangle + c_{1} | 1 \rangle.$$

Now I construct the density matrix (DM), $$\hat \rho = | \Psi \rangle \langle \Psi | = \frac{1}{2} \left( | 0 \rangle \langle 0 | + \exp( - \text{i} \phi )| 0 \rangle \langle 1 | + \exp( \text{i} \phi ) | 1 \rangle \langle 0 | + | 1 \rangle \langle 1 | \right).$$

So from the DM $\hat \rho$, I can read off $|c_{0}|^{2}$, $|c_{1}|^{2}$, $c_{0}c_{1}^{*}$, and $c_{0}^{*}c_{1}$. Basically $3$ equations and $4$ unknowns.

Is there a way to reconstruct $| \Psi \rangle$ uniquely from the DM, $\hat \rho$?

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    $\begingroup$ Uniquely up to the over-all phase, no? $\endgroup$ Apr 5, 2021 at 17:28
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    $\begingroup$ @CosmasZachos: Yes ofcourse. $\endgroup$
    – sbp
    Apr 5, 2021 at 17:46
  • $\begingroup$ Are you asking only for pure states and not including mixed states? $\endgroup$
    – TEH
    Apr 5, 2021 at 18:15
  • $\begingroup$ @TEH Yes as of now. $\endgroup$
    – sbp
    Apr 5, 2021 at 23:35

2 Answers 2

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On solving, we find:

$$\frac{1}{2}\begin{pmatrix} 1 & e^{-i\phi}\\ e^{i\phi} & 1 \end{pmatrix}= \begin{pmatrix} |c_0|^2 & c_0c^*_1\\ c_1c_0^* & |c_1|^2 \end{pmatrix} $$

$$\Rightarrow |c_0|=|c_1|=\frac{1}{\sqrt{2}}$$ $$c_0c^*_1=\frac{1}{2}e^{-i\phi}\Rightarrow c_0=e^{-i\phi}c_1$$ $$|\psi\rangle =c_0|0\rangle +c_1|1\rangle =c_0\left(|0\rangle+\frac{c_1}{c_0}|1\rangle \right)=\frac{1}{\sqrt{2}}e^{i\chi}(|0\rangle +e^{i\phi}|1\rangle )$$

So the wave function would be unique up to phase factor $\chi$.

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If you write $c_0 \equiv |c_0|\, e^{i\phi_0}$ and $c_1 \equiv |c_1|\, e^{i\phi_1}$, then you can write the wave function as

$$|\Psi\rangle = e^{i\phi_0}\left(|c_0| \,|0\rangle + |c_1|\,e^{i(\phi_1-\phi_0)}\,|1\rangle\right)\quad . $$

The associated density operator is then given by $\rho_{\Psi}\equiv |\Psi\rangle\langle \Psi|$. The diagonal elements will yield $|c_0|$ and $|c_1|$ and from the off-diagonal terms you can reconstruct $|c_1| \, e^{i(\phi_1-\phi_0)}$. However, you can only reconstruct the wave function up to the global phase, which is also intuitive, since two wave functions $|\Psi\rangle$ and $|\psi\rangle$ which differ only by a global phase will yield the same density operator.

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    $\begingroup$ Thats what I have thought as well. I can get $|c_{0}|^{2}$, $|c_{1}|^{2}$, and $\phi_{0}-\phi_{1}$ from $\hat \rho$. Ofcourse to some global phase it is underdertermined. $\endgroup$
    – sbp
    Apr 5, 2021 at 17:43

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