A two qubit state in a special form

Question

How can a pure two-qubit state $|\psi\rangle = a |00\rangle + b|01\rangle + c|10\rangle + d|11\rangle$, be written in the following form

\begin{equation} |\psi_{\alpha}\rangle = \sqrt{\alpha}|01\rangle + \sqrt{1-\alpha} |10\rangle. \end{equation} How can one prove this. And what would be $\alpha$ in terms of $a,~b,~c,~d$?

Answer 1

The trick is in the Schmidt decomposition - Following the theorem proof, we have two qubits, so two Hilbert spaces $H_1$ and $H_2$, both with dimension 2, with the bases defined as $\left\lbrace\left|e_0\right> = \left|0\right>_1,\, \left|e_1\right>=\left|1\right>_1 \right\rbrace$ and $\left\lbrace\left|f_0\right> = \left|0\right>_2,\, \left|f_1\right>=\left|1\right>_2 \right\rbrace$.

This defines a tensor $w = \sum_{i, j=0}^1\beta_{i, j}\left|i\right>_1\otimes\left|j\right>_2$. That is, $\beta_{00} = a, \,\beta_{01} = b, \,\beta_{10} = c$ and $\beta_{11} = d$.

We then write this as an $n\times n$ matrix $$M_w = \begin{pmatrix}\beta_{00} && \beta_{01} \\ \beta_{10} && \beta_{11}\end{pmatrix} = \begin{pmatrix}a && b \\ c && d\end{pmatrix}.$$

Now that we've got a matrix, we want to diagonalise it, so we perform a Singular Value Decomposition (SVD). That is, we want to write $M_w = UDV^\dagger$, where $D$ is a diagonal matrix (with the elements known as the 'singular values') and $U$ and $V$ and unitary $n\times n$ matrices. Or rather, to save having to do a chunk of maths, we know that the columns (also, rows) of both $U$ and $V$ each form an orthonormal basis - I'll call these $\left\lbrace\left|u_0\right>,\, \left|u_1\right>\right\rbrace$ and $\left\lbrace\left|v_0\right>,\, \left|v_1\right>\right\rbrace$.

Helpfully, the expression for the singular values of a $2\times2$ matrix is analytic: $$\sigma_{\pm} = \sqrt{\left|z_0\right|^2 + \left|z_1\right|^2 + \left|z_2\right|^2 + \left|z_3\right|^2 \pm \sqrt{\left(\left|z_0\right|^2 + \left|z_1\right|^2 + \left|z_2\right|^2 + \left|z_3\right|^2\right)^2 - \left|z_0^2 - z_1^2 - z_2^2 - z_3^2\right|^2}},$$ where \begin{align*}z_0 &= \frac{1}{2}\left(a+d\right) \\ z_1 &= \frac{1}{2}\left(b+c\right) \\ z_2 &= \frac{i}{2}\left(b-c\right) \\ z_3 &= \frac{1}{2}\left(a-d\right). \end{align*} This in turn gives $$\sigma_\pm= \sqrt{\frac{1}{2}\pm\sqrt{\frac{1}{4}-\left|ad - bc\right|^2}},$$ as a result of the normalisation condition $\left|a\right|^2 + \left|b\right|^2 + \left|c\right|^2 + \left|d\right|^2 = 1$. As $\sigma_+^2 + \sigma_-^2 = 1$, I'll redefine $\sigma_+ = \sqrt{\alpha}$ and $\sigma_- = \sqrt{\left(1-\alpha\right)}$, for reasons that should become clear below.

We can now write $M_w = \sqrt\alpha\left|u_0\rangle\langle v_0\right| + \sqrt{1-\alpha}\left|u_1\rangle\langle v_1\right|$.

'Rewriting' this as a tensor (as at the beginning¹) gives a state $$\left|\psi_\alpha\right> = \sqrt\alpha\left|u_0\right\rangle\otimes\left|v_0\right\rangle + \sqrt{1-\alpha}\left|u_1\right\rangle\otimes\left|v_1\right\rangle,$$ which is equivalent to what you have by defining $\left\lbrace\left|u_0\right> = \left|0\right>_u,\, \left|u_1\right>=\left|1\right>_u \right\rbrace$ and $\left\lbrace\left|v_0\right> = \left|1\right>_v,\, \left|v_1\right>=\left|0\right>_v \right\rbrace$, where $$\alpha= \frac{1}{2}+\sqrt{\frac{1}{4}-\left|ad - bc\right|^2}$$

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Calculating $M_w$:

$\left|u_0\right>$ is the left column of $U$ and $\left|u_1\right>$, the right column. Similarly, for $V$, $\left|v_0\right>$ is the left column and $\left|v_1\right>$, the right. This means that I can write $$U = \begin{pmatrix}\left|u_0\right> && \left|u_1\right>\end{pmatrix}$$ and $$V^{\dagger} = \begin{pmatrix}\left<v_0\right| \\ \left<v_1\right|\end{pmatrix}$$ so that $$M_w = UDV^\dagger = \begin{pmatrix}\left|u_0\right> && \left|u_1\right>\end{pmatrix}\begin{pmatrix}\sqrt\alpha && 0 \\ 0 && \sqrt{1-\alpha}\end{pmatrix}\begin{pmatrix}\left<v_0\right| \\ \left<v_1\right|\end{pmatrix},$$ which can be simplified as $M_w = \sqrt\alpha\left|u_0\rangle\langle v_0\right| + \sqrt{1-\alpha}\left|u_1\rangle\langle v_1\right|$.

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^{1 Even as a non-mathematician, I feel guilty just doing this}

Answer 2

Mithrandir's answer perfectly covers the functional details, but to state a clear response to the question...

It is not true that any two-qubit state $a|00\rangle+b|01\rangle+c|10\rangle+d|11\rangle$ can be written in the form $\sqrt{\alpha}|01\rangle+\sqrt{1-\alpha}|10\rangle$. Obviously, you can only get equality here if $a=0$ and $d=0$.

However, what you can do is a change of basis. An equivalent way of saying this is that there exists unitaries $U_A$ and $U_B$ such that $$ (U_A\otimes U_B)(a|00\rangle+b|01\rangle+c|10\rangle+d|11\rangle)=\sqrt{\alpha}|01\rangle+\sqrt{1-\alpha}|10\rangle $$

With that clarification, you could now follow Mithrandir's proof. I think of it in a different way (where you don't need to know what the singular value decomposition is, but you do need to know about the reduced density matrix). Consider the reduced density matrix of each way of writing the state, where we restrict to just the first qubit. We have $$ \rho_1=U_A\left((a|0\rangle+c|1\rangle)(a^\star\langle 0|+c^\star\langle 1|)+(b|0\rangle+d|1\rangle)(b^\star\langle 0|+d^\star\langle 1|)\right)U_A^\dagger\\=\alpha|0\rangle\langle 0|+(1-\alpha)|1\rangle\langle 1|. $$ Clearly, the right-hand side is diagonal, so $U_A$ must be the unitary that diagonalises the left-hand side. In other words, $\alpha$ and $1-\alpha$ are the eigenvalues of the matrix $$ \left(\begin{array}{cc} |a|^2+|b|^2 & ac^\star+bd^\star \\ a^\star c+b^\star d & |c|^2+|d|^2 \end{array}\right). $$ The $U_A$ always exists because that matrix is Hermitian. Similarly, $U_B$ is the unitary rotation that diagonalises the reduced density matrix on the second qubit.