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Displacement operator

  1. Jun 10, 2015 #1

    dyn

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    Hi . I've just encountered something called the displacement operator which is the exponential of a parameter multiplied by a vector but I thought the argument of an exponential had to be a scalar. Is this not true ?
     
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  3. Jun 10, 2015 #2

    DEvens

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    Provided it makes sense to apply the operator an arbitrary number of times you can take the exponential of the operator, at least in form.

    Remember the power expansion of exp(x) = 1 + x + x^2/2 + ... + x^n/n! + etc.

    So you can replace the 1 by the identity operator in the context. And the x by the operator. The x^2 means apply the operator twice, and so on.

    This does raise some thorny questions. For example, it is not trivial to know if this converges. If the operator has some nice Eigen values it may. But on the other hand, it may not. Also, it is not necessarily the case you can provide a nice closed form solution for exp(x) if x is an operator. And you may encounter lots of other worries with the expression.

    In quantum mechanics, one common thing you will encounter is the commutator of two operators.

    [x, y] = xy - yx

    If one of the operators is then put in exp, you have the interesting question of the meaning of such expressions as this.

    [exp(x),y] = exp(x) y - y exp(x)

    And that was the subject of a gnarly homework assignment in my first year of grad school.

    This also means you can put operators in some other functions, provided you have a power expansion for them. For example sin, cos, log, etc. Again, it does not always make sense, and there are plenty of complications.
     
  4. Jun 10, 2015 #3

    dyn

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    so in general the argument of the exponential should be a scalar but in certain circumstances such as this displacement operator the argument can be a vector ?
     
  5. Jun 10, 2015 #4

    Fredrik

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    You should provide some context. Where did you encounter this? Since you posted this in the differential geometry section, I'm thinking that you may be talking about the exponential map on a smooth manifold with a connection. http://en.wikipedia.org/wiki/Exponential_map_(Riemannian_geometry)
     
  6. Jun 10, 2015 #5

    dyn

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    I am self studying a relativity course. The displacement operator first came up in some notes on the Taylor series expansion of a scalar field along a curve.
     
  7. Jun 11, 2015 #6

    Fredrik

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    I think you will need to show us the notes.
     
  8. Jun 12, 2015 #7

    WWGD

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  9. Jun 12, 2015 #8

    dyn

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    To be honest functional calculus is over my head. I just thought the displacement operator would be more well known and I thought there would be a general rule whether exponentials could have a vector as an argument or not
     
  10. Jun 12, 2015 #9

    Fredrik

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    For the series expansion definition to make sense, the vector space must be equipped with a multiplication operation that's "nice enough" to ensure the convergence of the series. For example, it's sufficient that the space is a Banach algebra.

    The exponential map that I talked about is probably what you need, but I really can't tell since you still haven't posted a reference or explained what you meant by "displacement operator". Why haven't you? By not doing so, you have made it unreasonably hard to answer your question.
     
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