Sorry if this is a silly question, I am new to quantum computing
I was just reading this article that talked about the eigenstates of an operator. And I wonder, how can we find those eigenstates for a given operator? For example for the Hadamard gate, what are its eigenstates?
And also, why do we care about those eigenstates?
Thank you very much!
An eigenstate of an operator $U$ is a state $|v\rangle$ such that $U|v\rangle = c*|v\rangle$
Given a matrix $U$, the eigenvalues of $U$ are the values $\lambda \in \mathbb{C}$ such that $U |\psi \rangle = \lambda |\psi \rangle$. The state/vector $|\psi\rangle$ is the eigenstate/eigenvector of $U$. Note that we only care about the non-trivial case where $|\psi \rangle \neq \vec{0}$.
The Hadamard gate have the matrix representation of
$$ H = \dfrac{1}{\sqrt{2}} \begin{pmatrix} 1& 1\\ 1 & -1 \\ \end{pmatrix}$$
Its eigenvalues are the constants $\lambda \in \mathbb{C}$ such that
$$H|\psi\rangle = \lambda |\psi \rangle \hspace{1 cm} \Rightarrow \hspace{1 cm} \dfrac{1}{\sqrt{2}} \begin{pmatrix} 1& 1\\ 1 & -1 \\ \end{pmatrix} |\psi\rangle = \lambda |\psi \rangle $$
To find its eigenvalues and hence the corresponding eigenvectors, you need to solve the equations
$$\big| H - \lambda I \big| = 0$$ this is because $H|\psi\rangle = \lambda |\psi \rangle $ implies $H|\psi\rangle -\lambda |\psi \rangle = (H - \lambda I) |\psi \rangle = 0$. And this is only true if $\big| H - \lambda I \big| = 0$. Note that $|A|$ represents the Determinant of $A$.
Thus you are leaving with solving for
$$ \begin{vmatrix} \dfrac{1}{\sqrt{2} } - \lambda & \dfrac{1}{\sqrt{2} } \\ \dfrac{1}{\sqrt{2} } & -\dfrac{1}{\sqrt{2} } - \lambda \end{vmatrix} = 0 $$
This is equivalent to find the root of a polynomial (so not easy in general) but in this case you have a simple polynomial. You need to solve for
$$ \lambda^2 - 1 = 0 \hspace{1 cm} \Rightarrow \hspace{1 cm} \lambda = \pm 1$$
Thus the eigenvalues of $H$ is $\pm 1$.
Now you need to use these eigenvalues to determine the corresponding eigenvectors $|\psi \rangle$.
To do this, for example, taking $\lambda = 1$, then you can replace this back into the natrix
$$ \begin{pmatrix} \dfrac{1}{\sqrt{2} } - \lambda & \dfrac{1}{\sqrt{2} } \\ \dfrac{1}{\sqrt{2} } & -\dfrac{1}{\sqrt{2} } - \lambda \end{pmatrix} \Rightarrow \begin{pmatrix} \dfrac{1}{\sqrt{2} } - 1& \dfrac{1}{\sqrt{2} } \\ \dfrac{1}{\sqrt{2} } & -\dfrac{1}{\sqrt{2} } - 1\end{pmatrix}$$
Now, you are trying to find a vector $|\psi_1 \rangle = \begin{pmatrix} x \\ y \end{pmatrix}$ such that
$$\begin{pmatrix} \dfrac{1}{\sqrt{2} } - 1& \dfrac{1}{\sqrt{2} } \\ \dfrac{1}{\sqrt{2} } & -\dfrac{1}{\sqrt{2} } - 1\end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} $$
This is now just solving linear systems of equations! Upon doing this you will get that the eigenvector $|\psi_1 \rangle$ correspond to the eigenvalue $\lambda = 1$ is something like $|\psi_1 \rangle = \begin{pmatrix} 1+ \sqrt{2} \\ 1\end{pmatrix}$.
You can check this by seeing that
$$ H|\psi_1\rangle = 1\cdot |\psi_1\rangle \hspace{0.5 cm} \Rightarrow \hspace{0.5 cm} \dfrac{1}{\sqrt{2}} \begin{pmatrix} 1& 1\\ 1 & -1 \\ \end{pmatrix} \begin{pmatrix} 1+ \sqrt{2} \\ 1\end{pmatrix} = 1 \cdot \begin{pmatrix} 1+ \sqrt{2} \\ 1\end{pmatrix} $$
To find the eigenvector $|\psi_{-1}\rangle$ which corresponds to the eigenvalue $\lambda = -1$ you can follow the similar procedure.
