I have a question about the eigenvectors for the evolution operator of Grover's algorithm. Let $U=R_DR_f$, where $$\begin{align*} R_D &= 2|D\rangle\langle D| -I_N , \\ R_f &= I-2|x_0\rangle\langle x_0| \end{align*}$$ where $x_0$ is the (unique) marked element to find, and $|D\rangle$ is the uniform superposition of standard basis states.

I have three eigenvalues: $e^{\pm i(1-2/N)}$, and $-1$ with multiplicity $N-2$. My lectures say: Assuming that $x_0 = 0$, one eigenvector associated to $-1$ is $$|\alpha\rangle = \dfrac{|1\rangle-|2\rangle}{\sqrt{2}}.$$ I can verify this: $$\begin{align*} \Bigl(2|D\rangle\langle D| -I_N\Bigr)&\Bigl(I-2|0\rangle\langle 0|\Bigr)\left(\dfrac{|1\rangle-|2\rangle}{\sqrt{2}}\right) \\&= \Bigl(2|D\rangle\langle D| -I_N\Bigr)\left(\dfrac{|1\rangle-|2\rangle}{\sqrt{2}}-\dfrac{2|0\rangle \langle 0| 1\rangle}{\sqrt{2}}+\dfrac{2|0\rangle \langle 0| 2\rangle}{\sqrt{2}}\right) \\&= \Bigl(2|D\rangle\langle D| -I_N\Bigr)\left(\dfrac{|1\rangle-|2\rangle}{\sqrt{2}}\right) =-\dfrac{|1\rangle-|2\rangle}{\sqrt{2}}. \end{align*}$$

My question is: in the denominator of $|\alpha\rangle$, can we change the $\sqrt{2}$ to $\sqrt{y}$ for any $y\in \mathbb{R}$? I ask this because the particular scalar factor does not seem to affect the operation.


1 Answer 1


From the definition of an eigenvector, any eigenvector $|\psi\rangle$ of $U$ must satisfy $U |\psi\rangle = \lambda |\psi\rangle$ for some $\lambda$. Then, for any scalar factor $\alpha \ne 0$, we also have $$ U\Bigl(\alpha|\psi\rangle\Bigr) = \alpha \lambda |\psi\rangle = \lambda\Bigl(\alpha|\psi\rangle\Bigr)$$ so that $\alpha |\psi\rangle$ is an eigenvector as well, with the same eigenvalue.

So no, the scalar factor difference between $1/\sqrt 2$ and $1/\sqrt y$ is not important, as far as being an eigenvector goes. However, unless $y$ has complex modulus $1$, the second of these vector $\frac{1}{\sqrt y}\bigl(|1\rangle - |2\rangle\bigr)$ would not represent a normalised quantum state.


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