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Very basic questions - Hamiltonian

  1. Jun 26, 2008 #1
    After several failures in the past (why does the universe have to be so complicated?!), I'm once again trying to learn to understand the basics of QM, out of sheer frustration with not knowing what the heck physicists are talking about all the time. I know, I still have a long way to go.

    Anyway, I just started again and I'm already confused about something.

    I'm led to understand that the rules governing the time evolution of the quantum state, together with the definitions of the observable quantities (in the form of their associated operators) take the place of Newton's laws of motion and the classical definitions of the quantities. Is that about right?

    If so, why can't I find a list of the definitions of the operators of all the usual physics quantities written somewhere? I've found the operators for momentum, position, and spin, and they seem to make sense to me (with what little I know), but I can't find the definition of the Hamiltonian, which is needed for me to know how energy is defined in quantumland. More importantly, since the entire rules for how stuff happens are encoded in the Schrödinger equation, which relies on the undefined Hamiltonian, I can't imagine (or compute) how anything happens at all.

    Surely it has a definition somewhere? I mean, it can't just be that I make one up... thereby making up any laws of motion I want for my universe...

    Thanks, and apologies for my utter n00bity.
  2. jcsd
  3. Jun 26, 2008 #2


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    The energy of a particle in quantum mechanics is the same as in classical mechanics :

    [tex] H = \frac{p^2}{2m} + V(x) [/tex].

    Just replace the classical quantities with the respective operators, and you get the hamiltonian operator.
  4. Jun 26, 2008 #3
    All classical dynamical variables may be defined in terms of position and momentum. Therefore you have already found what you need.

    Regarding the Hamiltonian.....
    dx has shown you the general form of the Hamiltonian, where the first term is simply the kinetic energy (again, define in terms of momentum as mentioned above) and the second term is the potential. In a sense, you do get to "make up" your Hamiltonian. The potential is determined by you, God, and/or may be inherent to your system.

    For example, in the case of atomic hydrogen (one proton, one electron) the potential is simply the Coulomb potential. If you would like to set up your own crazy experiment with a charged particle placed off-center in an electric octupole, then you will have a more involved potential making for a more involved Hamiltonian.
  5. Jun 26, 2008 #4
    Thanks. But how/why can I replace quantities with operators? As I understood it, the relationship between a quantity and the operator is complicated and indirect: the quantity lives in the eigenvalues of the operator. I wouldn't expect to be able to add two operators in general and get the operator corresponding to a quantity that is the sum of the quantities represented by those two operators... would I? *brain smokes*

    And I guess the operator for that potential V(x) is just the operator that multiplies V(x) by phi(x) for all x... that appears to work out right.
  6. Jun 26, 2008 #5
    Not exactly sure what you mean by "quantities." Physical observables are represented by operators. All of your classical observables may be defined by position and momentum, so that's all you need (spin is another story, but that doesn't have a classical counterpart).

    What you might be getting tripped up on is the fact that operating the Hamiltonian on the wavefunction yields an eigenvalue equation. Well this is only true when you take out time dependence, hence the time-independent Schrodinger equation. Well, this is actually DERIVED from the more general time-dependent Schrodinger equation. From this derivation we see that the time-independent wavefunctions are actually energy eigenfunctions.

    The same does not hold true for either momentum or position. The reason for this is that you may not have a determinant state of either momentum or position (a la Heisenberg). So your eigenvalue equations will not exactly behave in the simple manner as we have discussed above.
  7. Jun 26, 2008 #6
    I believe that this is not something that can be proved through an arguement or derivation. I think that this is embedded in the postulates of quantum mechanics.
    Only if there is a corresponding classical equivalent.
    The x in V(x) is replaced with the operator X. This operator is defined such that X|[itex]\Psi[/itex]> = x|[itex]\Psi[/itex]>. V(X) is defined by its series expansion in the operator X.

  8. Jun 26, 2008 #7
    I was trying to put together another response, but I think my understanding of math is just not up to the task, which I think points to the fundamental problem. Maybe I still don't even understand the framework of the basic formulation of the theory. Can anyone recommend maybe what kinds of mathematics I should study to learn enough to approach basic QM? Any recommendations for URLs or, less desirably, books that might be a good first step?
  9. Jun 26, 2008 #8


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    You can study elementary QM using just the Schrodinger equation if you know some basic calculus and differential equations. If you want to study the full theory of QM you need to know at least advanced calculus and linear algebra (and also a good amount of physics to appreciate it fully).
  10. Jun 26, 2008 #9
    OK, well I think my vector cal is pretty good and my linear algebra is decent and improving, so I would think I should be able to understand this. But I guess I'm not getting the concept of an observable or an operator or something because so far it isn't making sense and I'm not even effectively communicating what the problem is.
  11. Jun 26, 2008 #10


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    If you're comfortable with vector calculus and linear algebra, you should be able to start learning quantum mechanics.

    In classical mechanics, systems have states which are points in a set. For example, a particle on a line has a position which is an element of R. The observables in classical mechanics are basically just the states, or numbers representing the states. In quantum mechanics, these points in sets are replaced by vectors in a vector space. The observables of quantum mechanics are not the states, but operators on this vector space. It's normal to feel a little uncomfortable with this at the beginning, but I suggest you not worry about this and go ahead and learn quantum mechanics. You can worry about what exactly it means and how you interpret it after you learn it.
  12. Jun 26, 2008 #11
    Wait, you're saying the observable somehow *is* the operator? I don't even get wat that means... I need a better book.
  13. Jun 26, 2008 #12
    You know what, never mind. I'm chalking this up as another abortive attempt. I'll wait a few years and try again. Sorry about this.
  14. Jun 26, 2008 #13


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    No I'm sorry, I wasn't clear. Let's consider a particle moving on a line. In classical mechanics, it is assumed that at every instant of time t, the particle has a certain position on the line x. That is it's state at time t. In quantum mechanics, the thing that is known at every instant of time is not its position, but something else. This is the state vector. This is a complex function defined on the line: [tex] |\psi(x)> [/tex]. That is the analog of the classical state.

    Now I will give a certain interpretation, which is not accepted by everyone. You can think of this state as somehow representing your knowledge about the particle. This is loosely just a probability density on the line. Now there are certain states which correspond to a certain knowledge that the particle is at a particular position. These are the eigenstates of the position operator [tex] \hat{X} [/tex]. In the case of a particle moving on the line, they are the Dirac delta functions. There are also momentum eigenstates, which are "states of knowledge" where the momentum is exactly known. These are the eigenvectors of the momentum operator which are

    [tex] e^{i \frac{p}{h}x}[/tex].

    If your "state of knowledge" is one of the momentum eigenstates, then it means that you know certainly what the momentum will be when you measure it.
  15. Jun 26, 2008 #14
    Given a classical observable f(q,p) there is a self adjoint operator F(Q,P) acting on Hilbert space that represents the expected outcomes.

    For simple cases
    [tex] H = \frac{p^2}{2m} + V(q) [/tex].
    this is easily understood. Unfortunately what to do for the
    [tex]q^np^m [/tex]
    terms is a little more complicated and not fully understood. A common formula is to transform
  16. Jun 26, 2008 #15


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    No! don't give up. Quantum mechanics is fascinating. Maybe you're just not reading the right books or something. I suggest the first few chapters of Feynman lectures vol 3 for an introduction. Also, Leonard Susskind (one of the founders of string theory) has a excellent set of video lectures on quantum theory you can find on youtube. Then try J. J. Sakurai's book if you are confident about your maths. Otherwise, try Shankar or Griffiths.
  17. Jun 27, 2008 #16


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    One of the postulates of QM is that states are represented by vectors in some vector space. A vector is often written using the "ket" notation: [itex]|\alpha\rangle[/itex]. If you describe the system as being in state [itex]|\alpha\rangle[/itex], then an observer who's translated in time relative to you would describe the system as being in another state [itex]|\alpha'\rangle[/itex]. There must be an operator U(t) that takes [itex]|\alpha\rangle[/itex] to [itex]|\alpha'\rangle[/itex], and it must be an exponential because it has to satisfy U(t+t')=U(t)U(t'), so we can write it as [itex]U(t)=e^{At}[/itex], where A is some operator. (Its existence is guaranteed by the condition U(t+t')=U(t)U(t')). It's convenient to choose U(t) to be unitary so that it doesn't change the norm of the state vectors it acts on. The requirement that U(t) be unitary implies that A must be anti-hermitian (i.e. [itex]A^\dagger=-A[/itex]). We prefer to work with hermitian operators because their eigenvalues are real, so let's define H=iA. H is now hermitian, and [itex]U(t)=e^{-iHt}[/itex].

    That's the definition of the Hamiltonian. It's the generator of translations in time. The momentum operators can be defined the same way as generators of translations in space, and the spin operators can be defined as generators of rotations.
  18. Jun 27, 2008 #17


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    Linear algebra is by far the most important kind, but you can probably just study those parts you need when you run into difficulties in your QM book. You obviously have to understand complex numbers too.

    Yes, an observable is an operator.
  19. Jun 27, 2008 #18
    OK, let's keep going I guess. Most of what's being said here are things I already understand, or at least think I understand. It's the details that are confusing me.

    Let me try stating the path I've followed and see if there are any objections. Maybe this will let me debug my brain glitch.

    1. The state of a particle is described by a set of complex numbers called a wavefunction. From one point of view, there can be said to be N complex numbers for each point in space, where N is the number of legal spin states. In this representation (the position/spin eigenstate basis) the probability of observing the particle with a particular combination of spin and position equals the squared magnitude of the complex number for that particular combination. Spin and position are therefore two quantities that the wavefunction tells us something about.

    2. The procedure for extracting information about other quantities (that is, quantities other than position and spin, which were very convenient in this basis) is more involved. To find out what the probability is of measuring the system to have a value p for a quantity (say, momentum), I have to solve the equation P Phi = p Phi, where Phi is an unknown vector of zillions of complex numbers and P is the operator corresponding to the quantity in question (momentum in this case). The result will be a wavefunction Phi that is the eigenstate of the system for the momentum value p. So now I take the inner product of Phi with the actual wavefunction and the result will be the probability of measuring the particle to have a momentum of p.

    *catches breath*

    3. Now that our wavefunction has some defined connection to measurable quantities, we have to find the rules for how wavefunctions behave. These rules are defined by saying that every wavefunction is a superposition of energy eigenstates (obviously). In each of those eigenstates, the components all keep the same magnitude with time, but their phases all spin around (counterclockwise!) at a frequency proportional to the value of energy for this eigenstate (that is, the eigenvalue).

    4. But wait! This doesn't help us yet! In order for that to have any bearing on what happens to the observable quantities in time, we need to know what the energy eigenstates look like in terms of the basis we were using at the beginning (the position/spin eigenstate basis), and that can only be done if we know the energy operator so we can solve an equation as in #2. So apparently the total energy operator is related to the momentum operator the same way that the energy quantity/eigenvalue/<insert jargon here> is related to the momentum <whatever>. And that is true because...?
  20. Jun 30, 2008 #19
    I have vague memories but...

    1) express the quantity/observable in terms of coordinates x, y, z and momentum [tex]p_x[/tex], [tex]p_y[/tex], [tex]p_z[/tex]

    2) the operator corresponding to a simple coordinate is the multiplication by such coordinate

    3) the operator corresponding to a momentum is [tex]-i \frac{h}{2\pi} \frac{\delta}{\delta x}[/tex]

    Therefore for example, if you need the operator for kinetic energy


    then you substitute each p with the corresponding

    [tex]-i \frac{h}{2\pi} \frac{\delta}{\delta x}[/tex],

    square it to

    [tex]-\frac{h^2}{4\pi^2} \frac{\delta^2}{\delta x^2}[/tex]

    and the final operator will be

    [tex]-\frac{h^2}{8m\pi^2} \nabla^2[/tex]
  21. Jun 30, 2008 #20
    Interesting. That derivation appears to implicitly assume that an operator applied twice corresponds to a quantity that is the square of the quantity represented by the original operator. Add that to the list of fascinating and unintuitive properties of quantum operators that have yet to be explained to me in any form I can understand. The other thing on the list right now is that adding two operators produces the operator corresponding to the quantity that is the sum of the quantites represented by the original two operators.

    If there is a trend here, it sounds like it would be that whatever you do to the operators will always "look the same" notationally as whatever happens to the quantities. I wonder how this mysterious connection is achieved, and how far it goes. I'm still thinking about it. Maybe this new information will help me come up with the answer.

    I'm trying to gather enough information that I can infer a full list of the basic postulates of QM. That way, I will be able to understand everything else based on how it is derived from those postulates. I don't know yet whether this particular question is related to some more postulates that I don't know, but it sounds interesting enough that it might be.
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