Changing Basis States for Two-Qubit States

In the Changing Quantum Basis States lesson, we learned how to express a general quantum state using a variety of basis states. That change of basis states is important in calculating the measurement outcome probabilities when the state preparation basis states are different from the measurement basis states. How does that work out for two-qubit states? You might also wonder if a change in basis states converts an entangled state into an unentangled state or vice versa.

In principle, the change-in-basis-states method is just what you might guess based on what we did in the Changing Quantum Basis States lesson. Let’s review the method. We are restricting ourselves to basis state changes that can be expressed in terms of a single angle $θ_{AB}$. In the “Complexifying Quantum States” chapter, we’ll show you what to do in more general cases.

We will write a single-qubit state in terms of two basis sets $\ket{A_0}$ and $\ket{A_1}$ and $\ket{B_0}$ and $\ket{B_1}$. We saw in the “Changing Quantum Basis States” lesson that we can write one set of basis states in terms of the other:

$$\begin{align*} \ket{A_0} &= \cos\theta_{AB}\ket{B_0} + \sin\theta_{AB}\ket{B_1}\\ \ket{A_1} &= -\sin\theta_{AB}\ket{B_0} + \cos\theta_{AB}\ket{B_1} \end{align*}$$

and

$$\begin{align*} \ket{B_0} &= \cos\theta_{AB}\ket{A_0} - \sin\theta_{AB}\ket{A_1}\\ \ket{B_1} &= \sin\theta_{AB}\ket{A_0} + \cos\theta_{AB}\ket{A_1}, \end{align*}$$

Here, $θ_{AB}$ is the state space angle between $|A_0⟩$ and $|B_0⟩$, and we assume that $|A_0⟩$ lies counterclockwise in state space by the angle $θ_{AB}$ from $|B_0⟩$. In a two-qubit state, we have products of the various individual qubit basis states. All we have to do is apply the one of the above two set of equations to the appropriate right-vector basis states.

Rather than writing a general result, which turns out to be rather messy algebraically, let’s do an example. Suppose we have two qubits $α$ and $β$ and we prepare them in a state

$$\ket\psi = \frac{1}{\sqrt2}(\ket{A_0}_\alpha\otimes\ket{A_0}_\beta + \ket{A_1}_\alpha\otimes\ket{A_1}_\beta).$$

The subscripts indicate which state vector applies to which qubit. We want to change the basis states for qubit α from the $A$ basis to the $B$ basis. Let’s break up the work into steps. First, we use the top line in the third-last equation in the first part inside the parentheses in the last equation:

$$\begin{align*} \ket{A_0}_\alpha\otimes\ket{A_0}_\beta &= \underbrace{[\cos\theta_{AB}\ket{B_0}_\alpha + \sin\theta_{AB}\ket{B_1}_\alpha]}_{\ket{A_0}_\alpha}\otimes\ket{A_0}_\beta\\ &= \ket{B_0}_\alpha\otimes(\cos\theta_{AB}\ket{A_0}_\beta)\otimes(\sin\theta_{AB}\ket{A_0}_\beta), \end{align*}$$

Here, we used $(a |F⟩) \otimes |G⟩ = |F⟩ \otimes (a |G⟩)$, valid for any number $a$.

We now do the same thing for the second term inside the parentheses in this equationEq_9_30 as follows:

$$\begin{align*} \ket{A_1}_\alpha\otimes\ket{A_1}_\beta &= \underbrace{[-\sin\theta_{AB}\ket{B_0}_\alpha + \cos\theta_{AB}\ket{B_1}_\alpha]}_{\ket{A_1}_\alpha}\otimes\ket{A_1}_\beta\\ &= \ket{B_0}_\alpha\otimes(-\sin\theta_{AB}\ket{A_1}_\beta)\otimes(\cos\theta_{AB}\ket{A_1}_\beta). \end{align*}$$

Finally, we add the last two equations and then group terms multiplying $|B_0⟩_α$ and $|B_1⟩_α$:

$$\begin{align*} \ket{A_0}_\alpha\otimes\ket{A_0}_\beta + \ket{A_1}_\alpha\otimes\ket{A_1}_\beta &= \ket{B_0}_\alpha\otimes\underbrace{(\cos\theta_{AB}\ket{A_0}_\beta-\sin\theta_{AB}\ket{A_1}_\beta)}_{\ket{B_0}_\beta} \\ &+ \ket{B_1}_\alpha\otimes\underbrace{(\sin\theta_{AB}\ket{A_0}_\beta+\cos\theta_{AB}\ket{A_1}_\beta)}_{\ket{B_1}_\beta}\\ &= \ket{B_0}_\alpha\otimes\ket{B_0}_\beta + \ket{B_1}_\alpha\otimes\ket{B_1}_\beta. \end{align*}$$

To get the final line in the above equation, we used this equationEq_9_29 for the $\beta$ qubit state.