Vector rotations and quantum mechanics [closed]

Quantum mechanics deals with wave function and complex numbers that can be seen as vectors in 2D plane. I am interested in vector rotations and there use in quantum mechanics.

What is the role that rotations play in quantum mechanics? Given a set of $N$ vectors in the complex plane, What is the quantum mechanical interpretation of rotating every vector by 45 degrees?

closed as unclear what you're asking by Michael Seifert, By Symmetry, Bill N, stafusa, Jon CusterJul 25 '18 at 1:07

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• What do you have in mind when you talk about "$N$ vectors in the complex plane"? Are you talking about the components of some state $|\psi \rangle$ in terms of some basis $|\phi_i \rangle$, or something else entirely? – Michael Seifert Jul 24 '18 at 13:29
• @MichaelSeifert I 'm not an expert in QM. Your free to interpret the N complex vectors in any appropriate way. – Mohammad Al-Turkistany Jul 24 '18 at 13:43
• I couldn't understand the actual question. A vector on the complex plane is actually just a complex number, $z=a+ib$. If you are asking about N complex numbers, than the rotation means to multiply them by a unit complex number, that is, $e^{i \theta}$ where $\theta$ is the angle. For the case 45 degrees (which means $\pi/4$ radians), it is $e^{i\pi/4}$ and it is equal to $i$. So for each $z$, you have $i\, z = -b+ia$. Is that your question? I think you are asking something more? If so, please clarify your question. – Oktay Doğangün Jul 24 '18 at 15:31
• @OktayDoğangün My question is about rotation as operator in quantum mechanics. What physical interpretation can be given to rotations in terms of information that can be extracted from quantum system? – Mohammad Al-Turkistany Jul 24 '18 at 16:25

There's quite a lot of physics behind this question. I will give a sloppy (but didactic, I hope) answer.

We know that we can use complex numbers to parametrize our states in a quantum mechanical system. For example, in position representation we have a wave function $\psi(x)$ that describes a system as the particle in a box or something like that. This wave function is complex-valued in general.

In order to extract physical information of the system we have, though, to compute expectation values of physical observables (hermitian operators) $H$. For example, the expected value of $H$ when the system is in the state $\psi$ is

$$\langle\psi|H|\psi\rangle = \int dx\ \psi(x)^*(H\psi(x))$$

The thing is that I can change $\psi(x)\mapsto e^{i\theta}\psi(x)$ (with $\theta$ a constant) and the result will be the same. So, if I am able to parametrize all the states of my system with functions $\psi_k$ this means that an equivalent one-to-one parametrisation in terms of physical information is $e^{i\theta}\psi_k$. This operation is basically rotating the space of states (the rotation in the complex-plane that the OP was talking about). It is a symmetry of the quantum mechanics, called phase invariance and what it tells us is that our physical description of the reality is redundant.

When one tries to generalize the QM to describe quantum fields, you want to get sure that this phase invariance occurs independently at every point in the space-time, so you promote the constant $\theta$ to a function of the coordinates $\theta(x)$ and then you get what is called gauge invariance, which is the most fundamental aspect in our formulation of the theories describing the elemental particles (fields).

Rotations play a very important role in quantum mechanics, but what is done is rotating the real space, $\mathbb{R}^3$, and then see what happens to wavefunctions.

I think you're asking a different thing, rotations of vectors themselves. Well, the thing is that vectors in QM live in a complex space $\mathbb{C}^n$. I don't know if you can, but I am completely unable to draw complex spaces except for $\mathbb{C}^1$; and if I can't draw them, I honestly "don't know" how to rotate complex vectors.

The only thing I can draw is $\mathbb{C}^1$. This can be, for example, a spin 0 particle, because that is $\mathcal{H}\otimes \mathbb{C}^1\sim \mathbb{C}^1$, and then you have a one-dimensional spinor, which depnds on the position!

$$\varphi(\vec{r})$$

If you rotate this complex number (in the complex plane), you are modifiying its value.

Rotations keep the norm, so it is still related to a probability density.

If you are turning all radii, it is a global unit phase factor, which is irrelevant.

Edit:

If you mean rotations in $\mathbb{R}^3$, it's a broad topic. It certainly deserves its own chapter, or even set of chapters. Let me summarise the key ideas:

• When we rotate a system in the real world, particles change position, so their probability to find them in the space has changed too.
• When the real systems rotates an angle $\theta$ about an axis given by a normal vector $\hat{n}$, wavefunctions also suffer a transformation given by an exponential of the angular momentum operator, which is obviously unitary:

$$U(\theta)=exp[(-i \theta (J_x n_x + J_y n_y + J_z n_z)]$$

• This is an infinite matrix impossible to compute, unless you restrict yourself up to a subspace. For example, a ½ spin particle preserves its ½ spin after transformations.
• In such cases, this exponential matrix can be calculated. However, in general, for the orbital part, we must make the transformation as

$$\varphi_F(\vec{x})=\varphi_0 (R^{-1} \vec{x})$$

which means "the wavefunction after rotation has the same values as the previous wavefunction had before rotation".

And then there are really interesting properties:

• If you rotate your measuring apparatus, you will be measuring another different magnitude, but it is related to the first one:

$$A_F=U A_0 U^{-1}$$

and the eignevalue will be obviously conserved.

• Operators that do not change under this transformations are scalar operators.
• Scalar operators are those that conmute with rotations, and hence with $\vec{L}$. A scalar hamiltonian conmutes with the angular momentum operator. Any spherically symmetric operator conmutes with $\vec{L}$
• That means that there exist a common basis of eigenvectors of both operators.
• The fact that $[H, \vec{L}]=0$ means that $\vec{L}$ is conserved through time passing.
• There is a natural degenerancy of energies due to this fact, as energies cannot depend on third components.

I hope this helps.

• I am fine with rotating the real space. – Mohammad Al-Turkistany Jul 24 '18 at 13:49
• Oh okay, I had understood you wanted to rotate vectors in the complex space. Please, see the edit for real space rotations. – FGSUZ Jul 24 '18 at 14:26