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Why is the vacuum normalised to 1?

  1. Apr 18, 2010 #1
    I'm studying (relativistic) quantum field theory and have a very simple question. Take for example the real scalar field. My notes simply state that <0|0>=1 is a very natural normalisation. If I'm not mistaken, then this implies that the probability of measuring the vacuum state is 1. How is this to be interpreted? If the field has an infinite number of states then will the probabilities add up to an infinite number? Does a probability greater than one correspond to the measurement of more than one particle? I think not because multiple particles correspond to a single state.

    Basically, I don't understand how the normalisation of states is to be interpreted in quantum field theory. Can anyone help?
     
  2. jcsd
  3. Apr 18, 2010 #2

    Matterwave

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    <0|0>=1 means that measuring the state |0> will yield the state |0> with probability 1 which is quite intuitive no?

    The probability of actually measuring the stat |0> on any arbitrary ket is <0|psi> which is not necessarily 1 no?

    This is just based off my knowledge from regular QM. If something changes in QFT then, disregard what I'm saying...
     
  4. Apr 18, 2010 #3
    I'm pretty sure you're right and I was just being stupid. At first I was thinking "shouldn't the probability that you measure nothing (the vacuum) be 0?". You'd think with this being my 4th quantum mechanics course I would know this. Oh well.
     
  5. Apr 18, 2010 #4

    Matterwave

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    Don't worry, these little confusions happen to the best of us. =)
     
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