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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!

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  • $\begingroup$ did you have a look at en.wikipedia.org/wiki/Eigenvalues_and_eigenvectors? There's plenty of information there. Could you pinpoint one aspect in particular you find unclear about it? $\endgroup$
    – glS
    Commented Apr 17, 2022 at 10:43
  • $\begingroup$ I will appreciate any example with Hadamard - what are its eigenvalues? And is any use case for eigenstates of H? $\endgroup$
    – Kasai
    Commented Apr 17, 2022 at 10:49
  • $\begingroup$ Welcome to QCSE. It is not a problem to ask simple questions here, they are also welcome. Check out here if you would like to understand how to ask good question quantumcomputing.meta.stackexchange.com/questions/370/…. Either way, welcome! $\endgroup$
    – R.W
    Commented Apr 17, 2022 at 16:05

1 Answer 1

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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.

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