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Basic question about QM Operators

  1. Jun 8, 2005 #1

    Just out of interest and curiosity, I am reading a book on QM which introduces QM from a highly practical point of view, minus the formalism, assuming basic knowledge of complex numbers, algebra, calculus and physics. I will probably do QM formally much later in college, but while going through the book, I came across some operations and I would like to confirm one such operation as it is not explicitly described in the book.

    Suppose we consider an experiment in which the energy of an electron in a hydrogen atom is measured N times. We denote the total energy in the i-th state by [itex]E_{i}[/itex] so that

    [tex]E_{i} = <i|\hat{H}|i>[/tex] (1)

    where [itex]\hat{H}[/itex] is the Hamiltonian operator. If [itex]C_{i}[/itex] for integers i, are complex numbers, then we since the system will be in some general state [itex]|\psi>[/itex], we can write it as a superposition of the eigenstates, i.e. as

    [tex]|\psi> = C_{1}|1> + C_{2}|2> + ..... + C_{i}|i>[/tex] (2)

    If [itex]n_{i}[/itex] is the number of times the system's energy is measured to be [itex]E_{i}[/itex] then, [itex]\sum_{i}n_{i} = N[/itex].

    First issue The book says that for large enough N, we must have

    [tex]\frac{n_{i}}{N} = |C_{i}|^2[/tex] (3)

    As I understand, the phrase "for large enough N" has either got something to do with statistics or with the fact that the measurement of energy forces the system to jump to one of the base states. Is this correct? Either way, is it correct to say that [itex]|C_{i}|^2[/itex] is a kind of weight which represents the fractional occurence of a particular energy [itex]E_{i}[/itex] in the distribution as the number of observations increases?

    Now, the average energy is given by

    [tex]E_{avg} = \frac{\sum_{i}n_{i}E_{i}}{\sum_{i}n_{i}} = \frac{\sum_{i}n_{i}E_{i}}{N}[/tex] (4)

    If N is large, we use equation (3) and write,

    [tex]E_{avg} = \sum_{i}|C_{i}|^{2}E_{i}[/tex]

    Second Issue

    Now, the book says that we use equation (1) and obtain,

    [tex]E_{avg} = <\psi|\hat{H}|\psi>[/tex] (5)

    This is the step I have trouble with. I tried to reason as follows. I know that

    [tex]<j|\hat{O}|i> = \int \psi^{*}_{j}O(x)\psi_{i}(x)dx[/tex]
    [tex]<j|\hat{H}|i> = \int \psi^{*}_{i}H\psi_{i}(x)dx[/tex]

    (where O is a dipole operator which causes the electron to go from state [itex]|i>[/itex] to state [itex]|j>[/itex]), because the book states these as more practical versions of the above relationships. So, I figured that we could write [itex]|C_{i}|^2[/itex] as

    [tex]|C_{i}|^2 = C_{i}^{*}C_{i}[/tex]

    and then use the definitions (2) and the conjugate properties of bra and ket states to get equation (5). I am not sure if this is correct because the trick (if at all it is valid) seems obvious from the integrals and from the commutativtity of the product of 2 complex numbers.

    I hope you will not mind mistakes or inadequacies in my reasoning/analysis too much as I have had virtually zero grounding in the development of QM in a rigorous mathematical fashion and I am pursuing it purely out of interest. I have done total derivatives, integrals, vectors, complex numbers, algebra and classical mechanics, electrodynamics minus tremendous partial derivatives (so no Lagrangian, Hamiltonian mechanics from the ab-initio).

    Thanks and cheers
    (PS--This is not homework, so I figured this is the right place for the post.)
    Last edited: Jun 8, 2005
  2. jcsd
  3. Jun 8, 2005 #2
    You have the correct idea for your first issue, the absolute square of the state-coefficients [tex]|C_{i}|^2[/tex] are interpreted as probability that state i is obtained after measurement and ni/N is just a statement that the law of large numbers works in probability theory eg. [tex]n_i/N[/tex] tends to the probability of [tex]n_i[/tex] when N goes to infinity.

    Edit: my post got all f*cked up cause some latex issue that displayed the wrong things.. so Ive edited away my strange interpretations :smile: if anyone happend to notice em.
    Last edited: Jun 8, 2005
  4. Jun 8, 2005 #3


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    Yes, it is the probability of measuring the energy at E_i

    [tex]E_{avg} = \sum_{i}|C_{i}|^{2}E_{i} = \sum_{i}|C_{i}|^{2} <i|\hat{H}|i> [/tex]

    But [tex]\sum_{i}C_i|i> = |\psi> [/tex] and the dual gives [tex]\sum_{i}<i|C_i^* = <\psi| [/tex]

    So plugging in gives the required result. Is this what you're saying ?
  5. Jun 8, 2005 #4


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    What books are you using?There are not too many books out there dealing with virtual statistical ensembles...

  6. Jun 8, 2005 #5


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    That's right. Just write [itex]|\psi \rangle = \sum_i C_i|i\rangle[/itex] and use the orthonormality of the eigenfunctions: [itex]\langle i|j\rangle = \delta_{ij}[/itex].
  7. Jun 8, 2005 #6
    Hi everyone

    Thanks very much for the clarification. Yes Gokul, thats what I meant.

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