Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Understand of vacuum expectation

  1. May 3, 2005 #1
    I have difficulty understanding the vacuum expectation:
    consider <0|A_{mu}|0>, we can understand it as the
    possibility ampitude of a photon turn into vacuum(although 0 in common),
    but in the spontaneous of gauge symmetry, we should understand
    <0|A_{mu}|0> as the strength of a electromagnetic field at vacuum.
    How can I arrive at this explaination?
  2. jcsd
  3. May 3, 2005 #2
    Calculate the ground state of a harmonic oscillator and yu will see it isn't zero.
  4. May 4, 2005 #3
    But what I mean is not the zero-point energy, but the field strength itself,
    for example, in electromagnetism the <state|A_{\mu}(x)|state>, does it mean the A value measured? or <state|E_{i}(x)|state>, does it mean the E value measured?
  5. May 4, 2005 #4


    User Avatar
    Science Advisor
    Homework Helper

    Generally speaking,those mathematical objects (called "transition probability amplitudes") are complex numbers,so they can't be measured in any way...Maybe their square modulus...:rolleyes:

  6. May 4, 2005 #5


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    That's the perturbative part: the quantisation of the small field excursions around their equilibrium value (which in QED, is taken to be the zero field).

    Well, in spontaneous symmetry breaking, the equilibrium value of the field is not zero. So the quantum particles (perturbatively) are the excursions around THAT equilibrium value ; the vacuum expectation value will then just give you both contributions: the equilibrium field value + plus the 1-quantum-to-zero-quantum transition amplitude. But as that last one is 0, you just get the equilibrium field value.

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?