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

M-<M> for M operator: why not a mismatch?

  1. Mar 19, 2014 #1

    nomadreid

    User Avatar
    Gold Member

    In "Quantum Computation and Quantum Information" by Nielsen & Chuang, on pp. 88-89, applying basic statistical definitions to operators, one of the intermediary steps uses the expression
    M-<M>
    where M is a Hermitian operator, and <M> is the expected value = <ψ|M|ψ> for a given vector ψ (that is, when one is testing for |ψ>.)
    What I do not understand is how one can subtract a vector from an operator. That is, <ψ|M|ψ> is a vector, and M is an operator. For example, if one took an example of M as a 2x2 matrix and ψ as a 1x2 vector, then <ψ|M|ψ> is a 1x2 vector, and then M-<M> has a mismatch in dimensions.
    What am I wrongly interpreting? Thanks.
     
  2. jcsd
  3. Mar 19, 2014 #2
    Usually, if ##A## is an operator and ##\lambda## is a number, we write ##A-\lambda##. This is of course nonsense, like you indicated. But what we mean by this is actually ##A-\lambda I##, where ##I## is the identity operator. So numbers should often be seen as multiplied with some identity operator that we don't write.
     
  4. Mar 19, 2014 #3

    nomadreid

    User Avatar
    Gold Member

    Thanks, micromass. That clears it up completely.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook




Loading...