When discussing the Dirac notation of an operator, for example, let's just say we have the bit flip gate $X$ if we want to write this in the Dirac notation does that just mean writing it as follows?



2 Answers 2


The Dirac notation for the Pauli-$X$ gate is:

$$|1\rangle \langle0| + |0\rangle \langle1|.$$

Now you might be wondering where this comes from. The term you're looking for is outer product representation of the $X$ gate. It follows from the spectral decomposition theorem (check Nielsen & Chuang 10th edition, p. 72) which holds for all normal operators. The key point:

In terms of the outer product representation, this means that $M$ can be written as $M=\sum_i\lambda_i|i\rangle\langle i|$,where $\lambda_i$ are the eigenvalues of $M$,$|i\rangle$ is an orthonormal basis for $V$, and each $|i\rangle$ an eigenvector of $M$ with eigenvalue $\lambda_i$.

The eigenvectors of the Pauli-$X$ gate are $-|0\rangle+|1\rangle$ and $|0\rangle+|1\rangle$, and the corresponding eigenvalues are $-1$ and $+1$ cf. Wolfram Alpha. Normalize the eigenvectors to get an orthonormal basis for $X$ i.e. $\{\frac{-|0\rangle+|1\rangle}{\sqrt{2}},\frac{|0\rangle+|1\rangle}{\sqrt{2}}\}$. According the spectral decomposition theorem you can represent the $X$ gate as:

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

To convince you that this result is correct let's apply it on an arbitrary qubit state $c_0|0\rangle+c_1 |1\rangle$:

$$(|1\rangle \langle0| + |0\rangle \langle1|)(c_0|0\rangle+c_1|1\rangle)$$ $$=c_0|1\rangle\langle0|0\rangle+c_1|0\rangle\langle 1|1\rangle$$ $$=c_0 |1\rangle + c_1 |0\rangle$$

So yes, our result is correct and the bits were indeed flipped upon application of $X=|1\rangle \langle0| + |0\rangle \langle1|$ to $c_0|0\rangle + c_1|1\rangle$.The last step followed from the fact that $\langle 0|0\rangle$ and $\langle 1|1\rangle$ are both equal to $1$, as $|0\rangle$ and $|1\rangle$ are orthonormal vectors i.e. their inner product $\langle \psi|\psi\rangle=1$ by definition.

We're done. As an exercise, find the outer product representation of the Pauli-$Z$ gate by yourself. And definitely, do go through the proof of the spectral theorem in Nielsen and Chung if time permits!


This might mean using the ketbra notation:

$$X = |1\rangle \langle0| + |0\rangle \langle1|$$

This notation describes the effect the operator has on the basis vectors: in this case $X$ converts $|0\rangle$ into $|1\rangle$ and vice versa.

A couple of other examples:

$$Z = |0\rangle \langle0| - |1\rangle \langle1|$$

$$\operatorname{CNOT} = |0 \rangle\langle0| \otimes I + |1 \rangle\langle 1| \otimes X = |00\rangle \langle00| + |01\rangle \langle01| + |11\rangle \langle10| + |10\rangle \langle11|$$

  • 1
    $\begingroup$ 0 and 1 in one of the outer products in X-gate should be flipped, I tried editing myself but says "Edits must be at least 6 characters; is there something else to improve in this post?" And I think everything else is perfect, P.S. The accepted answer has same typo, cheers! $\endgroup$
    – Hemant
    May 1, 2019 at 1:32
  • $\begingroup$ @Hemant Fixed, thank you! $\endgroup$ May 1, 2019 at 1:53

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