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Complex exponents

  1. Jan 14, 2007 #1

    daniel_i_l

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    I just started learning QM and have a question - lets say that i have the following amplitude for a particle:
    a*e^(-i*frequency*t)
    since the exponent doesn't change the probability, what is it's physical importance? from what i understood the frequency coorsponds to the energy - since energy is the tendency of the particle to stay in it's current state, a higher frequency would make the particle have more influence in the case of interference with another particle. so basically the exponent only has significance when it interacts with another particle?

    And one more thing, does the amplitude of the state change with time only when it's multiplied by a real exponent?
    Thanks.
     
  2. jcsd
  3. Jan 14, 2007 #2
    The state you describe is called a stationary state, or alternatively a pure state of, or even better an eigen state of, the Hamiltonian operator.

    Given an arbitrary wave function, it is not likely to be an eigenstate of the Hamiltonian for the configuration. But the eigenstates do form a basis in which we can express our arbitrary wave function as a linear combination of eigenstates of various (complex exponent) frequencies.

    Then the short answer is, the complex exponent is of critical importance when we have a mixed state, the normalized sum of two or more terms that you described above but having different frequencies.
     
  4. Jan 14, 2007 #3
    The term you cite is normally appended on the stationary wave function term that we get from a time independent Schrödinger equ. From this eq. the allowed frequencies are determined and the (normalized) wave functions of it give the physical information.
     
  5. Jan 14, 2007 #4

    CarlB

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    It's to define the momentum of the particle if that measurement were to be made.

    I'm sure they covered the "Heisenburg uncertainty principle". It's a restriction on how small you can make the product of the uncertainty in position and momentum. The absolute value of the wave function defines the probability distribution for a position measurement, but it says nothing about the momentum probability distribution.

    When you calculate the momentum probability distribution you will find that the imaginary stuff becomes important.
     
  6. Jan 15, 2007 #5

    jtbell

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    Staff: Mentor

     
    Last edited: Jan 15, 2007
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