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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


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    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


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    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.

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