I have a little difficulty with understanding. How do I properly visualize the change of qubit's basis as a rotation?

Let's say that we have classical basis vectors, $|0\rangle$ and $|1\rangle$. Now, we can change this basis into the $\{|+\rangle, |-\rangle\}$ one by applying a Hadamard gate:

$$ H|0\rangle = \frac{1}{\sqrt{2}}\begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix} \begin{bmatrix} 1 \\ 0 \end{bmatrix} = \frac{1}{\sqrt{2}} \begin{bmatrix} 1 \\ 1 \end{bmatrix} = \frac{|0\rangle + |1\rangle}{\sqrt{2}} = |+\rangle, $$

$$ H|1\rangle = \frac{1}{\sqrt{2}}\begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix} \begin{bmatrix} 0 \\ 1 \end{bmatrix} = \frac{1}{\sqrt{2}} \begin{bmatrix} 1 \\ -1 \end{bmatrix} = \frac{|0\rangle - |1\rangle}{\sqrt{2}} = |-\rangle. $$

Is there some analogy between the Hadamard gate and classical rotation matrix, which can be defined as

$$ R(\theta) = \begin{bmatrix} \cos \theta & -sin \theta \\ \sin \theta & \cos \theta \end{bmatrix},$$

for a rotation in the counterclockwise direction by the angle of $\theta$?

I think that my problem lies in viewing the vectors $|0\rangle$ and $|1\rangle$ as two perpendicular vectors on a plane of real numbers. But, we are dealing with two vectors of complex numbers, so I guess we shouldn't think of them in this way, but rather than that, use the Bloch sphere?

  • $\begingroup$ Might be interesting: a nice blog post about VQE, where basis changes are obtained via rotations around the X and Y axis in the Bloch sphere mustythoughts.com/post/…. $\endgroup$ – Davit Khachatryan Mar 14 at 12:22

Hadamard gate can be interpreted as a rotation in 3D Euclidean space (on Bloch sphere) by angle $\pi$ around X+Z axis.

The qubit rotation by angle $\theta$ around axis pointed by unit vector $\textbf{n}=\{n_x,n_y,n_z\}$ is described by rotation operator ($X$, $Y$ and $Z$ are Pauli matrices) \begin{align} R_{\textbf{n}}(\theta)=&n_xe^{-i\frac{\theta}{2}X}+n_ye^{-i\frac{\theta}{2}Y}+n_ze^{-i\frac{\theta}{2}Z}=\\ =&\begin{pmatrix} \cos\frac{\theta}{2}-in_z\sin\frac{\theta}{2}&(-in_x-n_y)\sin\frac{\theta}{2}\\ (-in_x+n_y)\sin\frac{\theta}{2}&\cos\frac{\theta}{2}+in_z\sin\frac{\theta}{2} \end{pmatrix} \end{align} If we choose $\theta = \pi$ and $\textbf{n}=\{\frac{1}{\sqrt{2}},0,\frac{1}{\sqrt{2}}\}$, then \begin{align} R_{\textbf{n}}(\pi)=&\begin{pmatrix} -\frac{i}{\sqrt{2}}&-\frac{i}{\sqrt{2}}\\ -\frac{i}{\sqrt{2}}&\frac{i}{\sqrt{2}} \end{pmatrix}=-\frac{i}{\sqrt{2}}\begin{pmatrix} 1&1\\ 1&-1 \end{pmatrix} \end{align}

which coincides with Hadamard gate up to the global phase $-i$

| improve this answer | |

The first postulate of quantum mechanics that can be found in the M. Nielsen and I. Chuang textbook:

Postulate 1: Associated to any isolated physical system is a complex vector space with the inner product (that is, a Hilbert space) known as the state space of the system. The system is completely described by its state vector, which is a unit vector in the system’s state space.

So, If one has some state $|\psi\rangle = \alpha |0\rangle + \alpha |1\rangle = \begin{pmatrix} \alpha \\ \beta \end{pmatrix}$, it can be described as a vector in the Hilbert space with $\alpha$ and $\beta$ complex coordinates. Because of normalization $|\alpha|^2 + |\beta|^2 = 1$, the length of that vector is $1$. When we apply a unitary operator the length doesn't change (still $|\alpha'|^2 + |\beta'|^2 = 1$). So unitary operations are just rotations of that vector in the Hilbert space (they don't change the length of it). If we change the basis with a unitary operator then it can be described as a rotation of basis vectors in the Hilbert space (the basis vectors will not change their lengths, they will just point to the other directions).

As mentioned in the comments, not only this. Like for rotations in the 3D Euclidean space, the inner product of two vectors in the Hilbert space is also preserving under the unitary transformation (Definition 2. or 3. from wiki). For 3D Euclidean space rotations, this also implies that the angles between the vectors are not changing. For basis vectors in the Hilbert space, this means that if they were orthogonal they will stay orthogonal after the unitary transformation.

| improve this answer | |
  • 1
    $\begingroup$ I would add that not only length but also angles among vectors are not changed under unitary transformation. $\endgroup$ – Martin Vesely Mar 14 at 15:42
  • 1
    $\begingroup$ Thanks, Martin, after your comment I changed the answer with proof that unitary operators are also preserving the inner product, but didn't have enough intuition to connect it with angles (or describe an angle) in the complex Hilbert space. If there are any ideas, you are more then welcome to add/change my answer. $\endgroup$ – Davit Khachatryan Mar 14 at 16:47
  • 1
    $\begingroup$ You are right. I was thinking about real vectors only. $\endgroup$ – Martin Vesely Mar 14 at 18:46
  • $\begingroup$ An update: instead of proving I added a reference to the alternative definitions of a unitary operator where the preservation of the inner product is included in order to have a shorter answer. $\endgroup$ – Davit Khachatryan Mar 14 at 19:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.