# Reconstructing wavefunction from the density matrix

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, $$\hat \rho$$?

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

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 $$\chi$$.

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.

• 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.
– sbp
Apr 5, 2021 at 17:43