Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Perpendicularity of quantuum states

  1. Jul 30, 2009 #1
    In quantum mechanics we learn that the states of a system described by eigen states which they are perpendecular. why should be this?
     
  2. jcsd
  3. Jul 30, 2009 #2
    Each eigenstate is an independent vector in the linear space of all different states, just like in linear algebra. Any arbitrary state can be represented as a linear superposition of such eigenstates (reper vectors). As soon as they are linearly independent, they can be orthogonal to each other.
     
  4. Jul 30, 2009 #3
    Well the state OF A SYSTEM is not necessarily orthonormal. It is a linear combination of orthonormal BASIS states. And basis states are orthonormal because it would make no sense to have a basis state which would have a non-zero projection onto another basis state. That would be like saying when I measure the momentum of a state that I know has momentum X I will find momentum X but also 30% of momentum Y.
     
  5. Jul 30, 2009 #4

    jtbell

    User Avatar

    Staff: Mentor

    A linguistic note: in English, we use the word "orthogonal" in this case, not "perpendicular".
     
  6. Jul 31, 2009 #5
    markoX -> You have to recall the general framework of QM to understand the answer. And in particular what you should keep in mind is that in QM the state of a system is described by some vector in a Hilbert space. And Hilbert spaces have the nice property of having a scalar product, which you can use to define some "orthogonality". (In other words, change the scalar product and the notion of orthogonality changes.) And it is using this scalar product that you can construct an "orthonormal basis" that you then use to describe every other vector in your Hilbert space. Think of [tex] \mathbb{R}^n[/tex]. When you write a vector in [tex] \mathbb{R}^n [/tex] as [tex](x_1,x_2,\ldots,x_n)[/tex] you are writing its components with respect to some particular basis. And it is simply a matter of convenience to pick this basis as an orthonormal basis. But you could have used another, non orthonormal, basis to describe the same vector. (Of course, then it's components would change.) And the situation in QM is completely analogous. It is simply mathematically convenient to work with an orthonormal basis, which you use to "decompose" your state vector. But you could, in principle, use a non orthonormal basis. Only your calculations would be much more complicated.

    maverick_starstrider -> What you say is a bit confused. You can always use a non orthonormal basis to decompose a generic vector. And such decomposition will always be unique, with no "overlapping projections". Think about the plane. As soon as you have two linearly independent vectors, you can write every other vector as a unique linear combination of the two given vectors. You can hence use them as your basis vectors. And the "projections" are then precisely the coefficients in the linear combination, and they are in no sense "overlapping".
     
  7. Jul 31, 2009 #6
    Yes but in quantum we always use orthonormal basis states. If they're not orthonormal you use gram-schmidt until they are.
     
  8. Jul 31, 2009 #7
    thanks for replies.
    I will keep in my mind linguistic note.
    yes that is right....It is esier to work with orthogonal basis.
    ok...thank again.
     
  9. Jul 31, 2009 #8
    Well it's not really linguistics so much as conceptual. Vectors in space can be said to be perpindicular because they are geometrical objects. States in quantum aren't geometrical objects, they aren't arrows in space.
     
  10. Jul 31, 2009 #9
    No, not always. Several electronic structure codes use non-orthogonal basis states because it's actually more convenient to not orthogonalize the basis states (the commonly used LMTO basis is one such nonorthogonal basis). If you want basis states represented by hydrogenic type orbitals for atoms in a molecule or solid, then basis states on different atoms won't be orthogonal. If you orthogonalize them, you wind up mixing different basis states so that your new basis states don't have the same physical meaning of localized orbitals.

    Also, Gram-Schmidt is not commonly used because it is not symmetric. A better approach is the Loewdin orthogonalization.
     
  11. Jul 31, 2009 #10
    I stand corrected.
     
  12. Jul 31, 2009 #11
    @markoX

    It's true, as many of the posters have mentioned, that the state of a system is represented by a vector in a Hilbert space, that the Hilbert space has many different bases, and that one and the same vector can be is the linear sum of many different basis vectors, and that there's nothing in the formalism of QM to pick out a particular preferred basis of orthogonormal vectors.

    (Though I think I perhaps have one slight caveat with what Bob_for_short wrote: it's not clear to me that just because there is linear independence there is also orthogonality. I believe that orthogonality involves a little more mathematical structure, namely an inner product, and this doesn't immediately come with the notion of linear independence)

    However, I notice that in your original question you particularly asked about *eigenstates*, and this notion introduces something new which may motivate the appearance of orthogonality.

    The Hilbert space, the space in which the vectors live, doesn't in and of itself contain anything which deserves to be called an eigenstate. Rather, eigenstates are connected to particular observables, such as energy, position or momentum; an eigenstate is always an eigenstate of some observable.

    Observables are represented in the Hilbert space formalism as *operators* on the Hilbert space: functions that map the vectors of Hilbert space onto other vectors of Hilbert space (analogous to operations such as reflection and rotation in Euclidean vector spaces). Given an operator, some vectors v in the Hilbert space have a particular property: v is mapped onto rv: a multiple of itself. These vectors are called the eigenstates of the operator, and r is an eigenvalue. Physically, the eigenvalue represents a possible value that the observable can have when measured.

    In quantum mechanics, the eigenstates that correspond to different eigenvalues are *orthogonal* to each other. This is not a convention, but something that follows from the mathematics of quantum mechanics and the way in which quantities are represented. This, then, is where the orthogonality comes from - but it is true that we do need to talk about observables before we it appears naturally, which is why the other answers here are perfectly correct too.

    On a more intuitive level, we perhaps can see why the different eigenvectors *should* be orthogonal as follows: Given an observable, say, Energy, its eigenstates correspond to all the ways in which the system would, with certainty, have a particular eigenvalue on measurement. (Of course, in QM, this is not normally the case, the typical state is a superposition or linear weighted sum of these eigenstates, these weights corresponding to the probability that, on measurement, it would have that particular eigenvalue.) Given an observable O, v2 one of its eigenstates, the inner product of two vectors v1 and v2 represents the probability that, given the system is in state v1, and a measurement is made, that the system will output the eigenvalue corresponding to v2. If v1 and v2 are two different eigenvectors of an observable corresponding to different eigenvalues, then the state v1 means that a measurement will yield the eigenvalue corresponding to v1 - so the probability of getting the eigenvalue corresponding to v2 *should* be zero. So the inner product should be zero, i.e. the vectors should be orthogonal.

    yossell

    (caveat: I'm no expert, so treat this with the healthy scepticism it probably deserves.)
     
  13. Jul 31, 2009 #12
    If v1 and v2 are two different eigenvectors of an observable corresponding to different eigenvalues, then the state v1 means that a measurement will yield the eigenvalue corresponding to v1 - so the probability of getting the eigenvalue corresponding to v2 *should* be zero. So the inner product should be zero, i.e. the vectors should be orthogonal.


    I think yossell is right , as quantum principle says a quantumic system consist of many discrete (or maybe continous ) states wich by measurment will project it to one of the states so if these states are not orthogonal it is'nt possible measure observable quantity (like energy , momentum , ... ) for an eigen state as yossell said .
     
    Last edited: Jul 31, 2009
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Perpendicularity of quantuum states
  1. Photon state (Replies: 1)

  2. Spin States (Replies: 9)

  3. Pointer states (Replies: 4)

  4. Preparing the state (Replies: 38)

Loading...