Can there be multiple energy eigenstates corresponding to the same eigenvalue of a Hamiltonian (Pauli-X)?

Question

all. I am a high-school student who has recently familiarized himself with linear algebra and is looking to understand quantum computing. So, I bought the classic textbook "Quantum Computation and Quantum Information" by Nielsen and Chuang.

In the book, I came across the problem: find the eigenvalues and eigenvectors of the Pauli matrices.

I started out with the Pauli-X matrix and correctly found its eigenvalues to be 1 and -1.

When I set about finding the eigenvectors (using the standard methods of linear algebra), however, I found that for an eigenvalue of 1, any scalar multiple of (1 1) would do, and for -1, it could be any scalar multiple of (-1 1).

So suppose that a qubit has a Hamiltonian of hωX (this is an example in the book). Its energy eigenvalues are hω and -hω, and its energy eigenstates are the same as the unit eigenvectors of X.

But based on the results of my above calculation, there are two options per eigenvalue. For hω they are $\frac{1}{\sqrt2}(|0\rangle + |1\rangle)$ and $\frac{1}{\sqrt2}(-|0\rangle -|1\rangle)$, and for -hω they are $\frac{1}{\sqrt2}(|0\rangle - |1\rangle)$ and $\frac{1}{\sqrt2}(-|0\rangle + |1\rangle)$. For each case, aren't these states distinct? Are they both right? The textbook only acknowledges the first state in each case.

Answer 1

The quantum states that differ by a global phase (i.e., by a complex number multiple which has absolute value of 1) are considered the same quantum state, since they can not be distinguished using any operations or measurements.

Thus, the eigenstates for hω are

• $|+\rangle = \frac{1}{\sqrt2}(|0\rangle + |1\rangle)$,
• $-|+\rangle = \frac{1}{\sqrt2}(-|0\rangle -|1\rangle)$,
• $i|+\rangle = \frac{1}{\sqrt2}(i|0\rangle + i|1\rangle)$
• and any other state you can obtain from $|+\rangle$ by multiplying it by a complex number and normalizing it.

The convention is usually to pick coefficients so that the state has a real positive number as the first coefficient.

Answer 2

For your more general question in the title, the answer is yes. This is more of a lesson in linear algebra than in quantum computing. What you found out is that actually the set of eigenvectors of a matrix $A$ corresponding to a given eigenvalue $\lambda$ forms a vector subspace of the space upon which $A$ acts, this is called the eigenspace of $\lambda$, $E_\lambda$. As a matter of fact, if $A v_1=\lambda v_1$ and $Av_2=\lambda v_2$, then $A(\alpha v_1+v_2)=\lambda(\alpha v_1+v_2)$, hence if $v_1,v_2\in E_\lambda$ then $\alpha v_1+v_2\in E_\lambda$ for all $\alpha \in \mathbb{C}$.

Now you might ask, are there even cases where there are two different vectors $v_1$ and $v_2$ that are in the same eigenspace? Well you found a trivial one, where the eigenspace is one dimensional, i.e.

$$ E_{\pm1}=\mathrm{span}\left(\frac{1}{\sqrt{2}}(|0\rangle\pm|1\rangle)\right)$$

they're one dimensional vector spaces, as as you noticed you can multiply the vector by any constant and it stays an eigenvector, but consider the matrix

$$ A=\begin{pmatrix} 0&1&0\\1&0&0\\0&0&1 \end{pmatrix}$$

you still have the eigenvalue $1$, but now it's different as

$$v_1=\begin{pmatrix}1\\1\\0 \end{pmatrix}v_2=\begin{pmatrix}0\\0\\1 \end{pmatrix} $$ are two linearly independent eigenvectors with the same eigenvalue, meaning that in this case

$$ E_1=\mathrm{span}(v_1,v_2)$$ the eigenspace is two dimensional, all linear combination of these two is an eigenvector with eigenvalue one. In this case we say that the eigenvalue is degenerate, specifically twofold degenerate or with degeneracy 2.

The most extreme example if of course the identity matrix, which has only one eigenvalue, $1$, and the eigenspace is the whole vector space.