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

Representing the state of a quantum system in different bases

  1. Oct 16, 2011 #1
    I am reading the book Introduction to Quantum Mechanics by David Griffiths and have come to the section on Dirac notation. It explains that the state of a quantum system is represented by a vector |β(t)> living out in Hilbert space, and, as with any vector, is independent of the choice of basis.

    It then explains that the position-space-wave-fuction [itex]\Psi(x,t)[/itex] is the coefficient in the expansion of |β(t)> in the basis of position eigenfuctions:

    [itex]\Psi(x,t)=<x|\beta(t)>[/itex] , where |x> is the eigenfunction of the position operator with eigenvalue x.

    The momentum-space-wave-function [itex]\Phi(p,t)[/itex] is the coefficient in the expansion of |β(t)> in the basis of momentum eigenfunctions:

    [itex]\Phi(p,t)=<p|\beta(t)>[/itex] , where |p> is the eigenfunction of the momentum operator with eigenvalue p.

    And similarly you can expand |β(t)> in the basis of energy eigenfunctions:

    [itex]c_{n}(t)=<n|\beta(t)>[/itex] , where |n> is the nth eigenfunction of the Hamiltonian (assumed discrete).

    So |β(t)> can be written in different way depending on the basis you choose. But I have a few questions which I'm hoping someone can clear up for me:

    1) What is |β(t)>? Am I correct in thinking that |β(t)> is [itex]\Psi(x,t)[/itex] when you're working in the position space, and |β(t)> is [itex]\Phi(p,t)[/itex] when you're working in the momentum space? So when I said at the start "|β(t)> lives out in Hilbert space", it's actually position Hilbert space or momentum Hilbert space, depending on which you're working with? And the different spaces will have different eigenfunctions for the operators?

    2) I am a bit confused by the dimension of |β(t)>. The eigenvalues of the position operator are always continuous (I think?), but say for instance the energy spectrum is discrete (or perhaps there are only a finite number of allowed energies). Are there not many more position eigenfunctions than energy eigenfunctions? Then when we write |β(t)> in each of these basis, will the dimensions not be different?

    I may have a few more questions but I'll leave it to those for the moment.

    Thanks for any help! :smile:
  2. jcsd
  3. Oct 16, 2011 #2
    No--there's just one Hilbert space. It's almost always good to make the analogy to the familiar vectors in R^3. You can write down the components of such a vector in any basis you want, but it's always the same vector and it always lives in the same space, R^3. It's possible to speak of vectors in an abstract way without ever writing down their components in any particular basis. I can talk about some vectors v and u, and their dot product v.u, without ever writing out the components of either. The vectors have an existence independent of their components in any particular basis. This abstract notion of a vector is what is meant by a ket. |β(t)> isn't a number, or a set of numbers, or a function; it's a Hilbert space vector independent of any wave function it might correspond to in a particular basis.

    Operators acting on Hilbert space vectors are just like matrices acting on regular vectors. They take a vector and do some linear transformation to it to produce another vector. For a certain class of matrices (and operators) it's possible to find eigenvectors that get taken to a multiple of themselves under the transformation such that these eigenvectors pick out a basis of R^3 (or Hilbert space) that it is natural to employ when discussing that matrix (operator). So the different operators determine different bases of the one Hilbert space, not different Hilbert spaces.

    Generally Hilbert space is infinite-dimensional, as you can tell from the fact that the various bases (position, momentum, energy) have infinitely many eigenvectors. To convince yourself that these all describe a space of the same size, you can represent any basis in terms of any other basis: you can write any momentum eigenstate as a linear combination of position eigenstates, and any position state as a linear combination of energy eigenstates, etc. So no basis describes more vectors than any other basis, since any state written in terms of one basis can be rewritten in terms of another.

    But you can also consider systems with a finite number of degrees of freedom, like a spin 1/2 particle with spatial degrees of freedom neglected, in which case the Hilbert space is finite dimensional. For a spin 1/2 particle we have 2 degrees of freedom and a 2-dimensional Hilbert space. This really is just a 2-dimensional vector space of the familiar sort; finite dimensional Hilbert spaces are useful to think about when you want to avoid the subtleties that can arise with infinite dimensional spaces.

    I'm not quite sure of the best way to address your concern about a discrete energy spectrum vs. the continuous position spectrum. But for realistic potentials that go to zero at infinity (unlike the ideal harmonic oscillator, say), there is always a continuum of energy eigenstates, possibly in addition to a number of discrete energy eigenstates representing bound states.
  4. Oct 17, 2011 #3
    Also I would like to add to the above that the physical interpretation of the state vector [itex]|\psi\rangle[/itex] still lacks a solidified interpretation, but Bell's theorem tells us some things that it cannot be.... There are a couple different interpretations about what the state vector means.
  5. Oct 17, 2011 #4
    [itex]\langle[/itex]Thanks for the responses guys. I'm still a little unclear on my first question though.

    Immediately after the section in the book which I quoted in my original post, the author states:

    "The functions [itex]\Psi[/itex] and [itex]\Phi[/itex], and the collection of coefficients [itex]{c_{n}}[/itex], contain exactly the same information - they are simply three different ways of describing the same vector:

    [itex]\Psi(x,t) = \int\Psi(y,t)\delta(x-y)dy = \int\Phi(p,t)\frac{1}{\sqrt(2\pi\hbar)}e^{ipx/\hbar}dp = \sum c_{n}e^{-iE_{n}t/\hbar}\psi_{n}(x)[/itex]" .

    So it's like he's saying that |β(t)> is [itex]\Psi(x,t)[/itex], and that this can be expressed in 3 different ways using either the position, momentum or hamiltonian operators eigenfunctions as bases. These eigenfunctions have the specific forms [itex]\delta(x-y)[/itex], [itex](1/\sqrt(2\pi\hbar)exp(ipx/\hbar)[/itex], etc, as shown above.

    But a few pages earlier (before we got to all this different bases stuff), he found the eigenfunctions for the momentum operator as [itex]f_{p}(x)=(1/\sqrt(2\pi\hbar)exp(ipx/\hbar)[/itex], and showed that these are complete, so we can write any function f(x) as

    [itex]f(x) = \frac{1}{\sqrt(2\pi\hbar)} \int c(p)e^{ipx/\hbar}dp[/itex].

    Therefore we can express [itex]\Psi(x,t)[/itex] in this form. The coefficient c(p,t) in this case is an important quantity which we call the momentum space wave function and give the symbol [itex]\Phi(p,t)[/itex]. Since the eigenfunctions are dirac orthonormal we can find [itex]\Phi[/itex] using Fourier's trick: [itex]\Phi(p,t) = <f_{p}|\Psi>[/itex]. Hence we have

    [itex]\Phi(p,t) = \frac{1}{\sqrt(2\pi\hbar)} \int e^{-ipx/\hbar}\Psi(x,t)dx[/itex] .

    There is then an exercise which I did to show that "in momentum space the position operator is [itex]i\hbar\frac{\partial}{\partial p}[/itex]", i.e. the expectation value of an observable Q(x,p) is

    [itex]<Q(x,p)> = \int \Phi^{*}\widehat{Q}(i\hbar\frac{\partial}{\partial p}, p)\Phi dp [/itex] in momentum space.

    So in "momentum space" the position eigenfunctions are now [itex]e^{-ipx/\hbar}[/itex]. But we already know that

    [itex]\Phi(p,t) = \frac{1}{\sqrt(2\pi\hbar)} \int e^{-ipx/\hbar}\Psi(x,t)dx[/itex]

    so we see that this is an expansion of [itex]\Phi(p,t)[/itex] in the basis of position eigenfunctions (in momentum space) with the coefficient being [itex]\Psi(x,t)[/itex]. But this ties in exactly with what I wrote in my original post:

    So it seems like if I expand |β(t)> in the basis of position eigenfunctions in position space then I can interpret |β(t)> as [itex]\Psi(x,t)[/itex], and if I expand |β(t)> in the basis of position eigenfunctions in momentum space then I can interpret |β(t)> as [itex]\Phi(x,t)[/itex]. In both cases we are in the basis of position eigenfunctions, but it seems like there are 2 different spaces?

    Thanks once again for your help! :-)

    P.S. Sorry if that was a bit long-winded but I just wanted to clearly explain my reasoning :tongue:
    Last edited: Oct 17, 2011
  6. Oct 17, 2011 #5


    User Avatar
    Science Advisor
    Homework Helper

    An analysis performed for example on the linear harmonic oscillatior reveals some things such as:

    * For each quantum system, if one assumes that the mathematical system describes one single system, one has a single rigged Hilbert space (RHS) in which its pure state representatives live.
    * Operators describing the quantum observables such as energy, momentum, position, angular momentum, electric charge, etc. act in this RHS (only one space) as self-adjoint operators, hence the same space (essentially the distributions space) admits different decompositions into a direct sum (or integral) of subspaces corresponding to the spectral values of the self-adjoint operator, as per the Gelfand-Kostyuchenko-Maurin theorem: we call such a decomposition a representation, and call them therefore: energy representation, position representation, number operator representation, spin representation, momentum representation, etc.

    In your example, the hamiltonian is assumed to have a pure point spectrum (for simplicity, I'm sure), i.e. a countable infinity of eigenstates. The position operator has an uncountable infinity of eigenstates, so yes, more.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook