Square root of CNOT and spectral decomposition of the Hadamard gate

Question

I'm trying to compute the spectral decomposition of the Hadamard gate but I'm making a mistake somewhere.

Note: I believe (though I may be wrong so correct me if I am) that spectral decomposition is a way to find a diagonalized version of a matrix, additionally I am trying to work out $\sqrt{\operatorname{CNOT}_{12}}$, and based off an exam paper question it said to do so knowing that $\operatorname{CNOT}_{12}=H_2Z_{12}H_2$. I know that $H_2=I \otimes H$, and I thought the best way to do this would be to diagonalize $H_2$ by diagonalizing $H$ then take the square root of the diagonals of $H_2, Z$ to find $\sqrt{\operatorname{CNOT}_{12}}$.

Say we have $$H=\begin{bmatrix} \tfrac{1}{\sqrt{2}} &\tfrac{1}{\sqrt{2}} \\ \tfrac{1}{\sqrt{2}} & -\tfrac{1}{\sqrt{2}} \end{bmatrix}.$$

Now spectral decomposition of this matrix will be $H=\sum_i \lambda_i|\psi_i\rangle \langle\psi_i|$, where $\lambda_i$ corresponds to an eigenvalue and $|\psi_i \rangle$ is its associated eigenvector.

First we find the eigenvalues:

$$\det(H-\lambda I)=\det \begin{bmatrix} \tfrac{1}{\sqrt{2}}-\lambda &\tfrac{1}{\sqrt{2}} \\ \tfrac{1}{\sqrt{2}} & -\tfrac{1}{\sqrt{2}}-\lambda \end{bmatrix}$$ $$=(\tfrac{1}{\sqrt{2}}-\lambda)(-\tfrac{1}{\sqrt{2}}-\lambda)-\tfrac{1}{2}$$ $$=-\tfrac{1}{2}+\lambda^2-\tfrac{1}{2}=-1+\lambda^2\implies \lambda=\pm 1$$

Now we find the eigenvectors:

$\lambda=1$:

$$\begin{bmatrix} x\\y \end{bmatrix}=\begin{bmatrix} \tfrac{1}{\sqrt{2}} &\tfrac{1}{\sqrt{2}} \\ \tfrac{1}{\sqrt{2}} & -\tfrac{1}{\sqrt{2}} \end{bmatrix}\begin{bmatrix} x\\y \end{bmatrix}$$

$$\implies \tfrac{x+y}{\sqrt{2}}=x \implies (\sqrt{2}-1)x=y$$

$$\tfrac{x-y}{\sqrt{2}}=y \implies (\sqrt{2}+1)y=x$$

These equations give eigenvectors $$v_1=\begin{bmatrix} 1 \\(\sqrt{2}-1) \end{bmatrix}, v_2=\begin{bmatrix} (\sqrt{2}+1)\\1 \end{bmatrix}$$

The eigenvectors for $\lambda=-1$ are found, similarly, to be $$v_3=\begin{bmatrix} 1 \\(-\sqrt{2}-1) \end{bmatrix}, v_4=\begin{bmatrix} (-\sqrt{2}+1)\\1 \end{bmatrix}$$ But $$H=-\begin{bmatrix} 1 \\(-\sqrt{2}-1) \end{bmatrix}\begin{bmatrix} 1 &(-\sqrt{2}-1) \end{bmatrix}- \begin{bmatrix} (-\sqrt{2}+1)\\1 \end{bmatrix}\begin{bmatrix} (-\sqrt{2}+1)&1 \end{bmatrix}+\begin{bmatrix} 1 \\(\sqrt{2}-1) \end{bmatrix}\begin{bmatrix} 1 &(\sqrt{2}-1) \end{bmatrix}+\begin{bmatrix} (\sqrt{2}+1)\\1 \end{bmatrix}\begin{bmatrix} (\sqrt{2}+1)&1 \end{bmatrix}$$

Doesn't give me a diagonal matrix, where have I gone wrong?

Answer 1

Firstly, there's a conceptual error in your calculation of the eigenvectors.

$$\begin{bmatrix} x\\y \end{bmatrix}=\begin{bmatrix} \tfrac{1}{\sqrt{2}} &\tfrac{1}{\sqrt{2}} \\ \tfrac{1}{\sqrt{2}} & -\tfrac{1}{\sqrt{2}} \end{bmatrix}\begin{bmatrix} x\\y \end{bmatrix}$$

$$(\sqrt{2}-1)x=y \tag{1}$$

$$(\sqrt{2}+1)y=x \tag{2}$$

(1) and (2) are equivalent equations. To convince yourself multiply both sides of (1) by $(\sqrt{2}+1)$. The point is that when you solve the eigenvalue equation, you do not get a single eigenvector but rather a linear subspace of eigenvectors. We can only put forth the form of the eigenvectors in that subspace as:

$$\begin{bmatrix}x \\ (\sqrt{2}-1)x\end{bmatrix}=x\begin{bmatrix}1 \\ (\sqrt{2}-1)\end{bmatrix}$$ where $x\in \Bbb C$.

Your other vector $\nu_2$ lies in this same linear subspace. It's equivalent to $(\sqrt{2}+1)\nu_1$.

And again, $\nu_4$ is simply $(-\sqrt{2}+1)\nu_3$.

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Secondly, for finding the square root of $\operatorname{CNOT}_{12}$ you could directly diagonalize $\operatorname{CNOT}_{12}$ instead of going about it in such a roundabout fashion (i.e. writing $\operatorname{CNOT}_{12}$ as $H_2Z_{12}H_2$). However, that approach is particularly useful for this specific case; see my update below.

It's a block diagonal matrix, so you can easily diagonalize the individual $2\times 2$ blocks.

Proceed like this:

Spectral decompose the upper left block like $P\Lambda P^{-1}$ and the lower right block like $Q\Lambda'Q^{-1}$. Then $$\operatorname{CNOT}_{12}=\begin{pmatrix}P&0\\0&Q\end{pmatrix}\begin{pmatrix}\Lambda&0\\0&\Lambda'\end{pmatrix}\begin{pmatrix}P^{-1}&0\\0&Q^{-1}\end{pmatrix}.$$

The required square root is simply $$\sqrt{\operatorname{CNOT}_{12}}=\begin{pmatrix}P&0\\0&Q\end{pmatrix}\begin{pmatrix}\sqrt\Lambda&0\\0&\sqrt\Lambda'\end{pmatrix}\begin{pmatrix}P^{-1}&0\\0&Q^{-1}\end{pmatrix}.$$

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Update:

Initially, I didn't understand the significance of writing the $\operatorname{CNOT}_{12}$ as $H_2Z_{12}H_2$ but the hint in the question paper helped. One obvious fact (as $H_2.H_2=\Bbb I$) is that $$\operatorname{CNOT}_{12}=H_2Z_{12}H_2 = (H_2\sqrt{Z_{12}}H_2)(H_2\sqrt{Z_{12}}H_2),$$ and so $$\sqrt{\operatorname{CNOT}_{12}}=H_2\sqrt{Z_{12}}H_2.$$

Note that $Z_{12}$ is the controlled phase gate for $\phi=\pi$.

Now, an interesting property of the controlled phase gate is:

$$\begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & e^{i\phi/2} \end{pmatrix}.\begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & e^{i\phi/2} \end{pmatrix}= \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & e^{i\phi} \end{pmatrix}$$

So the square root of the $Z_{12}$ gate i.e. $\sqrt{Z_{12}}$ is $$\begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & e^{i\pi/2} \end{pmatrix}.$$

Now just pre- and post-multiply $\sqrt{Z_{12}}$ with $H_2$'s and you're done! This is indeed simpler than the spectral decomposition method I suggested earlier.

Answer 2

There are two points where you are wrong.

First: When you compute the eigenvector of H, which is $2\times 2$ matrix. So there should be one eigenvector corresponding to an eigenvalue, and should be normalized. In your computation for the case of $\lambda=1$, if you make a normalization you will find that there are the same.

Second: If you just want to get the square root of a self-adjoint one, then just make a square root of the eigenvalue is fine. That is for a self-adjoint operator: $$A = \sum_i \lambda_i |\psi_i\rangle\langle\psi_i|$$ then $$\sqrt{A} = \sum_i \sqrt{\lambda_i}|\psi_i\rangle\langle\psi_i| $$