What is the spectral decomposition of the Pauli $X$ gate?

Question

The definition of spectral decomposition is as follows:

Assume the eigenvectors of $\hat{A}$ define a basis $\beta=\{|\psi_j\rangle\}$. Then $$A_{kj}=\langle\psi_k|\hat{A}|\psi_j\rangle=\alpha_j\delta_{jk},$$ and $$\hat{A}=\sum_{kj}A_{kj}|\psi_k\rangle\langle\psi_j|=\sum_{j}\alpha_{j}|\psi_j\rangle\langle\psi_j|.$$ But when I try to do this for the bit flip gate $X$, it doesn't seem to work. The eigenvectors are $|0\rangle+|1\rangle$ and $|0\rangle-|1\rangle$. Then if we take

\begin{align} X&=(\langle0|+\langle1|)X(|0\rangle+|1\rangle)\,\,\times\big[(|0\rangle+|1\rangle)(\langle0|+\langle1|)\big] \\ &+(\langle0|-\langle1|)X(|0\rangle-|1\rangle)\,\,\times\big[(|0\rangle-|1\rangle)(\langle0|-\langle1|)\big]. \end{align}

I don't get anything close to the $X$ matrix I was looking for. Could anyone please explain this to me?

Answer 1

The eigenvectors are $|\psi_1 \rangle = \frac{1}{\sqrt 2} (|0\rangle + |1\rangle)$ and $|\psi_2 \rangle = \frac{1}{\sqrt 2} (|0\rangle - |1\rangle)$ (remember to keep them normalized), with eigenvalues $a_1 = 1$ and $a_2 = -1$, respectively.

Now the spectral decomposition in terms of the outer product representation is:

$$X = a_1 |\psi_1 \rangle \langle \psi_1| + a_2 |\psi_2 \rangle \langle \psi_2| = $$

$$ = \frac{1}{2} (|0\rangle + |1\rangle)(\langle 0| + \langle 1|) - \frac{1}{2} (|0\rangle - |1\rangle)(\langle 0| - \langle 1|) = $$

$$ = \frac{1}{2} (\color{red}{|0\rangle\langle0|} + |0\rangle\langle1| + |1\rangle\langle0| + \color{red}{|1\rangle\langle1|}) - \frac{1}{2} (\color{red}{|0\rangle\langle0|} - |0\rangle\langle1| - |1\rangle\langle0| + \color{red}{|1\rangle\langle1|})$$

The red parts cancel out, so you're left with $X = |0\rangle\langle1| + |1\rangle\langle0|$.

Answer 2

Continue simplifying from what you've given

\begin{eqnarray*} X&=&(\langle0|+\langle1|)X(|0\rangle+|1\rangle)(|0\rangle+|1\rangle)(\langle0|+\langle1|)\\ &+&(\langle0|-\langle1|)X(|0\rangle-|1\rangle)(|0\rangle-|1\rangle)(\langle0|-\langle1|)\\ (\langle0|+\langle1|)X(|0\rangle+|1\rangle) &=& 0 + 1 + 1 + 0 = 2\\ (|0\rangle+|1\rangle)(\langle0|+\langle1|) &=& | 0 \rangle \langle 0 | + | 1 \rangle \langle 0 | + | 0 \rangle \langle 1 | + | 1 \rangle \langle 1 |\\ (\langle0|-\langle1|)X(|0\rangle-|1\rangle) &=& 0 - 1 - 1 + 0 = -2\\ (|0\rangle-|1\rangle)(\langle0|-\langle1|) &=& | 0 \rangle \langle 0 | - | 1 \rangle \langle 0 | - | 0 \rangle \langle 1 | + | 1 \rangle \langle 1 |\\ X &=& 2 (| 0 \rangle \langle 0 | + | 1 \rangle \langle 0 | + | 0 \rangle \langle 1 | + | 1 \rangle \langle 1 |) \\&+& (-2) ( | 0 \rangle \langle 0 | - | 1 \rangle \langle 0 | - | 0 \rangle \langle 1 | + | 1 \rangle \langle 1 |)\\ X &=& 4 | 1 \rangle \langle 0 | + 4 | 0 \rangle \langle 1 | \end{eqnarray*}

But there should be no factor of 4 there. So what was wrong? You didn't normalize your eigenvectors. That means you should divide all your vectors by $\sqrt{2}$.

In each of the 2 summands in your expression at the very top, there are 4 vectors. So each summand should be divided by $(\sqrt{2})^4=4$. Follow that through and the simplification works and you get $X=X$ rather than $X=4X$

Answer 3

TL;DR: You forgot to normalize the eigenvectors!

Your description of the spectral theorem is incomplete; it requires you to use an orthonormal basis i.e. the basis $\beta=\{|\psi_j\rangle\}$ must be orthonormal, as I previously discussed here. When normalized, the eigenvectors of the $X$ gate form the basis $\left\{{\frac{|0\rangle+|1\rangle}{\sqrt{2}}},{\frac{|0\rangle-|1\rangle}{\sqrt{2}}}\right\}$. Now let's evaluate the terms $\color{red}{\left(\frac{\langle 0|+\langle 1|}{\sqrt{2}}\right)X\left(\frac{|0\rangle+|1\rangle}{\sqrt{2}}\right)}$ and $\color{blue}{\left(\frac{\langle 0|-\langle 1|}{\sqrt{2}}\right)X\left(\frac{|0\rangle-|1\rangle}{\sqrt{2}}\right)}$. These are actually the eigenvalues of $X$ corresponding to the two eigenvectors, because if $|\psi\rangle$ is a normalized eigenvector (i.e. $\langle \psi|\psi\rangle = 1$ ) of $X$, and $\lambda$ is the corresponding eigenvalue, then $$X|\psi\rangle = \lambda |\psi\rangle \implies \langle \psi|X|\psi\rangle = \langle \psi|\lambda|\psi\rangle= \lambda\langle \psi|\psi\rangle =\lambda.1=\lambda.$$

Now you know that $X$ is the "bit-flip" gate, so it performs $|0\rangle \mapsto |1\rangle$ and $|1\rangle\mapsto|0\rangle$.

$$\therefore \color{red}{\left(\frac{\langle 0|+\langle 1|}{\sqrt{2}}\right)X\left(\frac{|0\rangle+|1\rangle}{\sqrt{2}}\right)} = \left(\frac{\langle 0|+\langle 1|}{\sqrt{2}}\right)\left(\frac{|1\rangle+|0\rangle}{\sqrt{2}}\right)$$ $$=\frac{1}{2}(\langle 0|1\rangle + \langle0|0\rangle+\langle 1|1\rangle + \langle 1|0\rangle)$$ $$=\frac{1}{2}(0+1+1+0)=\color{red}{1} \ (\because \text{$|0\rangle$ and $|1\rangle$ are orthonormal})$$

$$\therefore \color{blue}{\left(\frac{\langle 0|-\langle 1|}{\sqrt{2}}\right)X\left(\frac{|0\rangle-|1\rangle}{\sqrt{2}}\right)} = \left(\frac{\langle 0|-\langle 1|}{\sqrt{2}}\right)\left(\frac{|1\rangle-|0\rangle}{\sqrt{2}}\right)$$ $$=\frac{1}{2}(\langle 0|1\rangle - \langle0|0\rangle-\langle 1|1\rangle + \langle 1|0\rangle)$$ $$=\frac{1}{2}(0-1-1+0)=\color{blue}{-1} \ (\because \text{$|0\rangle$ and $|1\rangle$ are orthonormal})$$

So now according the spectral decomposition theorem you can represent the $X$ gate as:

$$\boxed{\color{blue}{-1}\left(\frac{|0\rangle-|1\rangle}{\sqrt{2}}\right)\left(\frac{\langle 0|-\langle1|}{\sqrt{2}}\right) + \color{red}{1}\left(\frac{|0\rangle+|1\rangle}{\sqrt{2}}\right)\left(\frac{\langle 0|+\langle1|}{\sqrt{2}}\right)}\tag{*}\label{*}$$ $$\require{cancel}=-\frac{1}{2}\left(\cancel{|0\rangle\langle0|}-|0\rangle\langle1|-|1\rangle\langle0|+\cancel{|1\rangle\langle1|}\right)+\frac{1}{2}\left(\cancel{|0\rangle\langle0|}+|0\rangle\langle1|+|1\rangle\langle0|+\cancel{|1\rangle\langle1|}\right)$$ $$=\color{green}{|1\rangle \langle0| + |0\rangle \langle1|}$$

To convince you that this result is correct let's apply it on an arbitrary qubit state $|\Psi\rangle=\alpha|0\rangle+\beta|1\rangle$:

$$\color{green}{(|1\rangle \langle0| + |0\rangle \langle1|)}(\alpha|0\rangle+\beta|1\rangle)$$ $$=\alpha|1\rangle\langle0|0\rangle+\beta|0\rangle\langle 1|1\rangle$$ $$=\alpha|1\rangle + \beta|0\rangle$$

So yes, the bits $|0\rangle$ and $|1\rangle$ of $|\Psi\rangle$ are flipped and our representation $\eqref{*}$ of the $X$ gate is indeed correct!