I need an explanation behind this calculation

Demon117
Messages
162
Reaction score
1
When expressing exp(\frac{i\sigma\cdot \widehat{n}\phi}{2}) as a series expansion, why can we make the assumption that (\sigma\cdot \widehat{n})^{2}=I?

Someone I asked on campus showed me this and I don't quite understand its implications (partly because I don't quite understand permutation tensors):

\sigma_{i}n_{i}\sigma_{j}n_{j}=n_{i}n_{j}(\delta_{ij}I+i\epsilon_{ijk}\sigma_{k})=(\widehat{n}\cdot \widehat{n})I+0=I

The middle two steps are really foreign to me, but I get the component parts in front and the last part. Could someone explain in further detail why the middle two parts exist?
 
Physics news on Phys.org
matumich26 said:
When expressing exp(\frac{i\sigma\cdot \widehat{n}\phi}{2}) as a series expansion, why can we make the assumption that (\sigma\cdot \widehat{n})^{2}=I?

Someone I asked on campus showed me this and I don't quite understand its implications (partly because I don't quite understand permutation tensors):

\sigma_{i}n_{i}\sigma_{j}n_{j}=n_{i}n_{j}(\delta_{ij}I+i\epsilon_{ijk}\sigma_{k})=(\widehat{n}\cdot \widehat{n})I+0=I

The middle two steps are really foreign to me, but I get the component parts in front and the last part. Could someone explain in further detail why the middle two parts exist?

Pauli matrices have the following IMPORTANT properties that you probably should verify and memorize:
\sigma_j^2 = I\,, [\sigma_i, \sigma_j] = 2 i\epsilon_{ijk} \sigma_k \,, \{\sigma_i, \sigma_j\} = 2\delta_{ij} I

Now, from the second step:
n_i n_j \sigma_i \sigma_j =<br /> \frac12 n_i n_j (\sigma_i \sigma_j + \sigma_j \sigma_i)<br /> = \frac12 n_i n_j \{\sigma_i, \sigma_j\} = n_i n_j \delta_i I = I

What the person you used is this:
\frac12 \left( [\sigma_i, \sigma_j] + \{\sigma_i, \sigma_j\} \right) = \delta_{ij} I + i\epsilon_{ijk} \sigma_k
 
Last edited:
mathfeel said:
Pauli matrices have the following IMPORTANT properties that you probably should verify and memorize:
\sigma_j^2 = I\,, [\sigma_i, \sigma_j] = 2 i\epsilon_{ijk} \sigma_k \,, \{\sigma_i, \sigma_j\} = 2\delta_{ij} I

Now, from the second step:
n_i n_j \sigma_i \sigma_j =<br /> \frac12 n_i n_j (\sigma_i \sigma_j + \sigma_j \sigma_i)<br /> = \frac12 n_i n_j \{\sigma_i, \sigma_j\} = n_i n_j \delta_i I = I

What the person you used is this:
\frac12 \left( [\sigma_i, \sigma_j] + \{\sigma_i, \sigma_j\} \right) = \delta_{ij} I + i\epsilon_{ijk} \sigma_k

Wow, that is actually very helpful. Thank you. But I wonder, is the last portion \frac12 \left( [\sigma_i, \sigma_j] + \{\sigma_i, \sigma_j\} \right) = \delta_{ij} I + i\epsilon_{ijk} \sigma_k necessary? It looks as though you can get by without that from the second step. Thanks again.
 
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. Towards the end of the first lecture for the Qiskit Global Summer School 2025, Foundations of Quantum Mechanics, Olivia Lanes (Global Lead, Content and Education IBM) stated... Source: https://www.physicsforums.com/insights/quantum-entanglement-is-a-kinematic-fact-not-a-dynamical-effect/ by @RUTA
Back
Top