# How to calculate the state vector of a mixed state of two superposition state?

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 $$\left|\psi\right\rangle$$ ?

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.

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