Let $\left|\psi_1\right\rangle=\alpha_1\left|0\right\rangle+\beta_1\left|1\right\rangle$ , and $\psi_2=\alpha_2\left|0\right\rangle+\beta_2\left|1\right\rangle$.

Assume that $\left|\psi\right\rangle$ is a mixed state of $w_1*\left|\psi_1\right\rangle$ and $w_2*\left|\psi_2\right\rangle$ with $w_1+w_2=1$.

How to calculate the state vector of $\left|\psi\right\rangle$ ?


1 Answer 1


If the state is a mixed state of $|\psi_{1}\rangle$ and $|\psi_{2}\rangle$ (or any two states) it cannot be written as a state vector; state vectors only describe pure states, as opposed to mixed states.

If it is a coherent superposition of $|\psi_{1}\rangle$ and $|\psi_{2}\rangle$, you can write:

$$\sqrt{w_{1}}|\psi_{1}\rangle + \sqrt{w_{2}}|\psi_{2}\rangle = (\sqrt{w_{1}}\alpha_{1} + \sqrt{w_{2}}\alpha_{2})|0\rangle + (\sqrt{w_{1}}\beta_{1} + \sqrt{w_{2}}\beta_{2})|1\rangle$$

The implicit assumption here is that $|\psi_{1}\rangle$ and $|\psi_{2}\rangle$ are orthogonal. If this is not the case, it makes less sense to sum the two states in such a way, but one can always normalize using their inner product $1+\sqrt{w_{1}w_{2}}|\langle\psi_{1}|\psi_{2}\rangle|^{2}$ (as you already pointed out in your comment):

$$ \frac{1}{1+\sqrt{w_{1}w_{2}}|\langle\psi_{1}|\psi_{2}\rangle|^{2}}\Big( \sqrt{w_{1}}|\psi_{1}\rangle + \sqrt{w_{2}}|\psi_{2}\rangle\Big) $$

If it is not a coherent superposition but indeed a (statistical) mixture of the two, a good (but wieldy) description of the state is:

"I am not certain what the state of the qubit is. It is either $|\psi_{1}\rangle$ or $|\psi_{2}\rangle$, but not a proper superposition of the two. With probability $w_{1}$ it is $|\psi_{1}\rangle$, and with probability $w_{2}$ it is $|\psi_{2}\rangle$."

Now, that is quite a mouthful and therefore we also have a different technique of writing down the state, making use of density matrices. The density matrix $\rho$ describing the state is

(with $\langle\psi_{1}|$ the complex transpose of $|\psi_{1}\rangle$):

$$ \begin{split} \rho &= w_{1}|\psi_{1}\rangle \langle\psi_{1}| + w_{2}|\psi_{2}\rangle \langle\psi_{2}| \\ &= w_{1} \begin{bmatrix}\alpha_{1}\alpha^{*}_{1} & \alpha_{1}\beta^{*}_{1} \\ \alpha^{*}_{1}\beta_{1} & \beta_{1}\beta_{1}^{*}\end{bmatrix} + w_{2} \begin{bmatrix}\alpha_{2}\alpha^{*}_{2} & \alpha_{2}\beta^{*}_{2} \\ \alpha^{*}_{2}\beta_{2} & \beta_{2}\beta_{2}^{*}\end{bmatrix} \\ \end{split} $$

For more info you could check, for instance, the Wikipedia page on density matrices or a good Quantum mechanics or quantum computing book.

  • $\begingroup$ Thank you. If it is a coherent superposition, why is the vector of the mixed state not met the normalization constraint? $(w_1\alpha_1+w_2\alpha_2)^2+(w_1\beta_1+w_2\beta_2)^2\ne 1$ $\endgroup$
    – Devymex
    Dec 5, 2020 at 14:52
  • $\begingroup$ Oops, my bad - that should've been $\sqrt{w_1}$ and $\sqrt{w_2}$ $\endgroup$
    – JSdJ
    Dec 5, 2020 at 14:54
  • $\begingroup$ But still $(\sqrt{w_1}\alpha_1+\sqrt{w_2}\alpha_2)^2+(\sqrt{w_1}\beta_1+\sqrt{w_2}\beta_2)^2=w_1+w_2+2\sqrt{w_1w_2}\left(\alpha_1\alpha_2+\beta_1\beta_2\right)\ne 1$, if $\alpha_1\alpha_2+\beta_1\beta_2\ne0$ $\endgroup$
    – Devymex
    Dec 5, 2020 at 15:03
  • $\begingroup$ See updated text $\endgroup$
    – JSdJ
    Dec 5, 2020 at 15:19

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