A single particle quantum state [tex]\Psi[/tex] is given a by a wave function (vector) representable as a linear superposition of eigenvectors [tex]\Upsilon[/tex] weighted by a number proportional to their probability of occurrence:(adsbygoogle = window.adsbygoogle || []).push({});

[tex]\Psi[/tex]=a_1 [tex]\Upsilon_1[/tex]+a_2 [tex]\Upsilon_2[/tex]+.....

In the case of the observable "spin", there are only two eigenvectors, which constitute a 2 dimensional basis.

In the case of the observable energy or momentum, there are more than 2 eigenvectors (n-dimensional basis)

What determines which basis is used in representing the state [tex]\Psi[/tex]?

Is it just a matter or choice, based on which observable we want to know?

Also, if [tex]\Psi[/tex] is represented in with two different basis sets, then the components of the sets must be derivable from each other....but it seems hard to be able to express all the energy, momentum, position eigenvector in terms of only the two spin eigenvectors...

thanks!

**Physics Forums - The Fusion of Science and Community**

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Quantum state: how many bases?

Loading...

Similar Threads - Quantum state many | Date |
---|---|

A Can disjoint states be relevant for the same quantum system? | Mar 13, 2018 |

A States in usual QM/QFT and in the algebraic approach | Feb 14, 2018 |

What's the energy-spread of the quantum Universe state? | Feb 13, 2018 |

Conceptual question of many particle quantum systems and state vectors | Jan 8, 2014 |

What was the quantum state at t=0 (big bang). Many thanks. | Feb 26, 2011 |

**Physics Forums - The Fusion of Science and Community**