Physical interpretation of the rotation of a ket

Click For Summary
SUMMARY

The discussion centers on the physical interpretation of rotating a quantum ket, specifically in the context of spin-1/2 particles. The rotation operator for spin-1/2 about the z-axis is defined as R(φ) = exp(-i S_z φ/ħ), with φ representing the angle of rotation. An example is provided using the state |α⟩ = (1/√2)(1, 1) as the +−eigenstate of Sx, which, when rotated by 90 degrees (φ = π/2), transforms into the +−eigenstate of Sy. The resulting state after rotation is |α_R⟩ = (e^(-iπ/4)/√2)(1, i), confirming the expected transformation.

PREREQUISITES
  • Understanding of quantum mechanics concepts, particularly quantum states and kets.
  • Familiarity with rotation operators in quantum mechanics.
  • Knowledge of spin-1/2 systems and their eigenstates.
  • Basic proficiency in linear algebra, specifically matrix operations.
NEXT STEPS
  • Study the mathematical formulation of rotation operators in quantum mechanics.
  • Learn about the implications of quantum state transformations under various operators.
  • Explore the concept of eigenstates and eigenvalues in the context of quantum spin systems.
  • Investigate the physical significance of quantum state rotations in experimental setups.
USEFUL FOR

Quantum physicists, students of quantum mechanics, and researchers interested in the behavior of spin-1/2 particles and their transformations.

bjnartowt
Messages
265
Reaction score
3
physical interpretation of the "rotation" of a ket

Homework Statement



No problem statement. I'm just having trouble imagining what it means to "rotate" a quantum -ket, especially since not all -kets are eigenstates of position. I know what the rotation operator is. I also can picture rotating a coordinate system in my head--that's easy of course. But what does it mean when a -ket is rotated?


Homework Equations



see attached .pdf. they are notes I've taken from shankar p. 306-306.

The Attempt at a Solution



see attached .pdf. they are notes I've taken from shankar p. 306-306.
 

Attachments

Physics news on Phys.org


Take the spin of an electron, for example. Suppose it's in the state
\vert \alpha \rangle = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ 1 \end{pmatrix}
which is the +-eigenstate of Sx. If you rotate it by 90 degrees about the z-axis, you'd expect it to be in the +-eigenstate of Sy. The rotation operator about the z-axis for spin-1/2 is
R(\phi)=\exp\left(-\frac{i S_z \phi}{\hbar}\right) = \begin{pmatrix}e^{-i\phi/2} & 0 \\ 0 & e^{i\phi/2}\end{pmatrix}
For \phi=\pi/2, you get
\vert \alpha_R \rangle = R(\pi/2)\vert \alpha \rangle = \frac{1}{\sqrt{2}} \begin{pmatrix}e^{-i\pi/4} \\ e^{i\pi/4}\end{pmatrix} = \frac{e^{-i\pi/4}}{\sqrt{2}} \begin{pmatrix} 1 \\ i \end{pmatrix}
which is indeed the +-eigenstate of Sy.
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
3K
Replies
8
Views
3K
  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 16 ·
Replies
16
Views
4K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 5 ·
Replies
5
Views
11K
  • · Replies 4 ·
Replies
4
Views
7K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K