# What are the eigenstates of an operator?

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!

• 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?
– glS
Apr 17, 2022 at 10:43
• I will appreciate any example with Hadamard - what are its eigenvalues? And is any use case for eigenstates of H? Apr 17, 2022 at 10:49
• 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!
– R.W
Apr 17, 2022 at 16:05

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.