Ladder operators

Discussion in 'Physics & Math' started by QuarkHead, Oct 27, 2010.

  1. QuarkHead Remedial Math Student Valued Senior Member

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    Also known as creation/annihilation operators.

    It seems these guys were introduced by Paul Dirac to simplify the solution to certain eigenvalue-eigenvector equations in quantum physics.

    These operators are, as far as I can see, usually handed to us on a plate, and take on slightly different forms, depending on who you read. But I think there is a logic here, but I also have some questions of mine own.

    And just to satisfy Guest, I will confess this is just as much a sharing of personal revelation as a question.

    Consider the quantum harmonic oscillator in one dimension whose Hamiltonian may be written as \(H=\frac{1}{2}\omega^2 q^2 + \frac{1}{2}p^2\), where the \(p,\,\,q\) are respectively momentum and position operators, and \(\omega\) is a constant that may loosely be taken as the natural resonance frequency of the system. (Note I have not included a mass term; this doesn't matter as it drops out in all subsequent calculations).

    So let me attempt a "factorization" of \(H\). That is, treat \(p,\,\,q\) as though they were simple vector-valued variables, not operators. I find that \(\frac{\omega q +ip}{\sqrt{2\omega}},\,\,\,\,\,\frac{\omega q -ip}{\sqrt{2\omega}} \) factorize \(H\) up to a multiplicative constant. (Notice these two are Hermitian conjugates)

    In other words, by defining \(a = \frac{\omega q +ip}{\sqrt{2\omega}}\) and \(a^{\dag}=\frac{\omega q -ip}{\sqrt{2\omega}}\) I easily find, by simple rearrangement, that \(H = \omega(aa^{\dag})= \omega(a^{\dag}a)\).

    Now this is a cheat, of course, since the mixed terms in my expansion should not cancel (the momentum and position operators don't commute!). But setting \(\hbar =1\) and \(p = -i\frac{\partial}{\partial x}\) and \(x=x\) (I can justify this if requested, albeit in a hand wave way), and using the commutator \([p,q] = pq-qp=i \) I find that

    \(aa^{\dag} = \frac{H-\frac{1}{2}\omega}{\omega}\) so that \(H= \omega(aa^{\dag} +\frac{1}{2})\).

    Wow. By being honest about my non-commuting operators, all I have done is added a constant \(\frac{\omega}{2}\) as compared to when I was "cheating". I find this a little surprising. Should I? Well perhaps not; I have done the calculations and they are good. But I am still a little surprised.

    Having done the calculation and looking at the definition of my ladder operators, I can see by inspection that I will have that \(H =\omega(a^{\dag}a -\frac{1}{2})\) which immediately tells me how to express\(p\) and \(q\) in terms of these operators. Maybe later, but for now let me get to the point.

    I said these operators are usually handed on a plate. So I pulled another definition of these ladder operators at random off the internet. Here:

    \(a= \sqrt{\frac{\omega}{2}}q + \frac{i}{\sqrt{2 \omega}}p\) and similarly for its Hermitian conjugate \(a^{\dag} = \sqrt{\frac{\omega}{2}}q - \frac{i}{\sqrt{2 \omega}}p\). These are clearly related to the definitions I gave above, but I cannot make them algebraically identical.

    But the amazing thing is that all the equalities above still hold. So do the expressions for \(p,\,\,q\) (which I haven't yet convinced you I have extracted). I am still surprised. Again should I be?

    My guess is no: any definition of these operators will suffice provided only that they satisfy the above equalities. But it is just that - a guess

    So my main question is this: Do there exist situations where I should prefer to use one definition of the ladder operators over another?
     
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  3. alephnull you can count on me Registered Senior Member

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    Am I missing a trick here?
    Your definition: \( a = \frac{\omega q +ip}{\sqrt{2\omega}} = \frac{\omega}{\sqrt{2 \omega}} q + \frac{i}{\sqrt{2\omega}}p \)

    Multiplying top and bottom by root omega ("rationalising the surd") we get

    \( a = \frac{\sqrt{\omega}\omega}{\sqrt{2 \omega^2}} q + \frac{i}{\sqrt{2\omega}}p = \frac{\sqrt{\omega}\omega}{\sqrt{2} \omega} q + \frac{i}{\sqrt{2\omega}}p = \sqrt{\frac{\omega}{2}} q + \frac{i}{\sqrt{2\omega}}p \)

    This is exactly the form given by the definition from the internet.
     
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  5. temur man of no words Registered Senior Member

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    They are identical!
     
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  7. rpenner Fully Wired Valued Senior Member

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    I see four lights!
     
  8. QuarkHead Remedial Math Student Valued Senior Member

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    *blush*
     
  9. rpenner Fully Wired Valued Senior Member

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    Point #2.
    But:
    \(\omega a a^\dag = \omega (\frac{\omega q +ip}{\sqrt{2\omega}})(\frac{\omega q -ip}{\sqrt{2\omega}}) = \frac{1}{2} (\omega^2 q +p^2 + i\omega pq - i\omega pq) = H + \frac{i \omega}{2} [ p,\, q ] = H - \frac{\omega}{2}\)
    and
    \(\omega a^\dag a = \omega (\frac{\omega q -ip}{\sqrt{2\omega}}) (\frac{\omega q +ip}{\sqrt{2\omega}})= \frac{1}{2} (\omega^2 q +p^2 - i\omega pq + i\omega pq) = H - \frac{i \omega}{2} [ p,\, q ] = H + \frac{\omega}{2}\)
    and presumably, \(\omega \neq 0\)
     
  10. prometheus viva voce! Registered Senior Member

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    On a slightly more serious and interesting note, the important thing about ladder operators is that one can take the vacuum and generate states by repeated action with the creation operator. Schematically: \(\left(a^\dag (\omega_1) \right)^{n_1} \left(a^\dag (\omega_2) \right)^{n_2} \ldots| 0 \rangle = |n_1,n_2, \ldots\rangle\), ie we have a state containing \(n_1\) particles of energy \(\omega_1\) etc. Similarly, when one acts with the annihilation operator it reduces the particle number, unless there are no particles with that \(\omega\), and in that case we get zero: \( a(\omega_i) |\ldots 0_i \ldots \rangle = 0\).

    Now suppose we have a situation where we have 2 vacuum states. That may sound a little loopy, but an example of this situation is where you have flat space with 2 observers, one of which is inertial and the other moves with a constant acceleration. Here there will be 2 sets of ladder operators (say a and b) to correspond to the two vacuum states \(|0_a \rangle\) and \(|0_b \rangle\) (also assume for simplicity that we can only create particles with 1 energy \(\omega\)). With this set up, suppose we are in a's vacuum state and we want to know how many particles are observed (that sounds like a tautology, but indulge me). We do this by sandwiching the particle number operator \(N = a^\dag a\) with the state so obviously \(\langle 0_a | a^\dag a |0_a \rangle = 0\) since \(a |0_a \rangle = 0\), but according to observer b, the particle number operator is \(N = b^\dag b\) and the number of particles observed is \(\langle 0_a | b^\dag b |0_a \rangle\) so it's not clear without doing the explicit calculation whether this is zero or not.

    In the setup described, it's not zero and one can show that if the inertial observer sees a vacuum the accelerating observer sees a thermal spectrum. It's called the Unruh effect and is very closely related to the emission of particles by black holes.
     
  11. QuarkHead Remedial Math Student Valued Senior Member

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    I don't understand your point, rpenner, sorry.

    What you quoted of mine was me "pretending" that \(p,\,\,q\) were arbitrary variables rather than non-commuting operators.

    But what you calculated is exactly as I found when I act like a grown-up and accepted they are operators. My contraction \(H = \omega(aa^{\dag}+\frac{1}{2})\) (which I did state) follows instantly from your calculation. Etc....

    Look, I as good as held my hand up to a foolish schoolboy error (thanks to alephnull and temur for that), don't try to make me look even more foolish
     
  12. AlphaNumeric Fully ionized Registered Senior Member

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    Nice reference

    Please Register or Log in to view the hidden image!

     
  13. Guest254 Valued Senior Member

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    Come on guys! I've just googled "I see four lights", and the result doesn't sway far from the archetypal "internet physics board joke". Shame on you all.
     
  14. AlphaNumeric Fully ionized Registered Senior Member

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    It's a Star Trek reference. Picard gets kidnapped and subjected to torture where the interrogator turns on a set of lights and asks over and over the simple question "How many lights do you see?". There's 4 but the interrogator keeps saying "How can you be mistaken, there's 5." and tortures Picard every time he says there's 4 lights. He'll know he's broken Picard when Picard says there's 5 lights. Finally when Picard gets released his final words are a screamed "there .... are .... FOUR .... lights!!". Uncyclopedia once had it as one of their spoofs of the Wikipedia main page 'Did you know....?' factlets.

    /edit

    http://www.youtube.com/watch?v=moX3z2RJAV8
     
  15. Guest254 Valued Senior Member

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    I just died a little inside!
     
  16. CptBork Valued Senior Member

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    Just some comments about ladder operators for those interested, on top of what prometheus said:

    Similar techniques can be used to solve other problems in quantum physics. I remember using it to describe the time evolution and energy eigenstates for particles such as neutrons passing through a magnetic field (did it to solve a homework problem, and got punished with reduced marks for not solving it using a method that was only introduced and developed in a later chapter). I've also seen ladder operators used in a similar fashion to find the energy eigenstates for the non-perturbative hydrogen atom.

    Basically you can solve all of these problems, including the simple harmonic oscillator, by just plugging them straight into the Schrodinger equation and crunching through the math. But the cool part is that in certain cases, you can take advantage of the algebraic properties of the operators used to describe these systems, and from those algebraic properties alone you can deduce the full solution without resorting to partial differential equations. I strongly recommend you look at the topics of spin quantization and eventually Lorentz group representation theory, because in addition to being fundamentally important topics in quantum physics, these subjects make use of techniques very similar to the ladder operators you've already discussed here.
     
  17. Green Destiny Banned Banned

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    Can't you see the relevence of her post? It's quite obvious she is referring to creation and annihilation operators. For instance you can substitute for the ladder operators by working backwards on the matrix counterparts of the creation and annihilation operators obtained from the quantum harmonic oscillator model. http://en.wikipedia.org/wiki/Creation_and_annihilation_operators
     
  18. QuarkHead Remedial Math Student Valued Senior Member

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    Hmm.. please explain, in simple terms, what you mean, and most especially the phrase "working backwards on the matrix counterparts". This is Greek to me. Obviously you understand it, and I confess I don't. Please help me

    But I thank you for alerting me to the existence of Wikipedia - I would never have found it without your help
     
  19. Green Destiny Banned Banned

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    You really need to start putting two and two together. I linked it because I extracted the information from there, because I recognized Rpenner was referring to the creation and annihilation operators.
     
  20. alephnull you can count on me Registered Senior Member

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    I just assumed that rpenner was pointing out the fact that quarkhead said \( \omega aa^{\dagger} = H \) when actually using the definition of a given, \( \omega a a^{\dagger} = H - \frac{ \omega }{2} \)
    But quarkhead adressed this point.

    Your post is a bit odd.
    Of course they're referring to the creation/annihilation operators (quarkhead even states this in the first line of the OP), what you say after that about matrices is irrelevant and kinda weird.

    I have a question about the OP:

    What do you mean by justify? Aren't these just the definitions of the position and momentum operators in 1 dimension?
     
  21. Green Destiny Banned Banned

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    Oh so he knew? I thought he was unaware Rpenner was referring to the operators. I guess my post would look kind of odd in face of that.
     
  22. AlphaNumeric Fully ionized Registered Senior Member

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    I would say its pretty obvious GD did a Google search and is just parroting back the Wiki link, he doesn't know how to actually do said quantum mechanics.

    'Extracted' being your phrase for "Parroting without understanding".
     
  23. Green Destiny Banned Banned

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    Well, what have you.
     

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