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

Eigenstates, Eigenblahs, and Eigenpoofs? HELP O.O

  1. Jun 10, 2008 #1
    Okay, so here's my knowledge of quantum mechanics: *crickets chirping in the background*. Here's my mathematical knowledge: basic integral calculus, some multivariable calculus, decent amount of linear algebra (knowledge of subspaces, projections, eigenvalues of matrices, eigenvectors, etc.), a decent amount of diff. eqs (not PDEs, though), all of high school math (extremely well... I was like 3 points of the usamo this year).

    So anyways, I started learning quantum mechanics like a week ago, and pretty much have learned:

    the construction of the Schrodinger equation (time dependant and independant)
    the overall concept of a Hermitian operator
    the infinite well situation

    and that's about it.

    I have been running into the words: eigenvalues, eigenfunctions, and eigenstates many times. I am pretty sure I understand what an eigenvalue and eigenfunction is, but I'm a bit shaky on what an eigenstate is. Just to clarify, in the equation

    [tex]
    \hat{A}\psi = a \psi
    [/tex]

    for some [tex]a[/tex], we have that [tex]\psi[/tex] is an eigenfunction (which could be vector-valued?) of [tex]\hat{A}[/tex] and [tex]a[/tex] is an eigenvalue of [tex]\hat{A}[/tex]. Is this correct?

    If relevant or helpful at all, I am currently a senior in high school who is going to enter college as a freshman this coming fall.
     
  2. jcsd
  3. Jun 10, 2008 #2

    Fredrik

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Yes, it is. The "which could be vector-valued" comment might be an indication that you have misunderstood one thing though. If you define the sum of two functions by (f+g)(x)=f(x)+g(x) and the product of a number and a function by (af)(x)=af(x), then you have turned the set of all functions into a vector space. My point is that it's not the values that the functions take that are vectors. The functions are the vectors, according to the definition of a vector space. The functions you encounter in quantum mechanics are going to be complex-valued functions of 1-3 real variables.
     
  4. Jun 10, 2008 #3
    Yes, by this I meant that [tex]\psi[/tex] takes in a complex valued vector and outputs another complex valued vector... each component of the input is fed into the respective component of [tex]\psi[/tex].
     
  5. Jun 11, 2008 #4

    Fredrik

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    The "input" is real-valued, and the "output" is just one complex number. The input represents a position in space and time. So I shouldn't have said 1-3 variables. It's 4 variables unless we ignore some of the spatial dimensions and/or the time dependence. [itex]\psi:\mathbb{R}^4\rightarrow\mathbb{C}[/itex]

    When your book starts talking about spin, you will probably see expressions like [itex]\psi(\vec x)\chi[/itex], where [itex]\psi[/itex] is the same as before and [itex]\chi[/itex] is a [itex]n\times 1[/itex] matrix (n, the number of rows, depends on the spin of the particle).
     
  6. Jun 12, 2008 #5

    vanesch

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    If I may add something, trying to de-confuse :smile: the OP.

    In quantum mechanics, the state of a system (be it a particle, a molecule, or anything else we want to handle with quantum mechanics) is described by, well, a quantum state. For single particles, that quantum state corresponds to "a wavefunction", that is, a complex-valued function over space: psi(x,y,z). Historically, Schroedinger thought it was going to be something similar to the electric potential, V(x,y,z): a kind of field in space. But later, people realized that what was important, was the fact that in the set of quantum states, the superposition principle holds: if A is a state, and B is a state, then a A + b B must also be a state. Turns out that that works pretty well for "wavefunctions": if psi1(x,y,z) is a quantum state, and psi2(x,y,z) is a quantum state, then psi3(x,y,z) = a psi1(x,y,z) + b psi2(x,y,z) is also a quantum state, a wavefunction. But it will turn out that quantum states are not always just "functions over x,y and z" ; it is just in the single particle case that this is so. What is important, is that superposition principle holds, which means that any quantum system must have a set of quantum states which forms a vector space.

    It is from that realization that we call the set of quantum states, the "state space" and a quantum state also a "state vector". In the case of a single particle, the state vector is nothing else but a wavefunction of the form psi(x,y,z).

    The more "sophisticated" quantum mechanics one does, the less one thinks of "wave functions", and the more one thinks of "vectors in a vectorspace".

    Note that the vector space of wavefunctions psi(x,y,z) is infinite-dimensional.

    In fact, the vector space of quantum states will turn out to have some other mathematical properties (apart from just being a vector space), and mathematically, a set obeying these properties is called a Hilbert space.
     
  7. Jun 12, 2008 #6
    Wouldn't it be fair to add that for [itex]N[/itex] particle systems, we have a wave function of the form
    [tex]\psi:\mathbb{R}^{3N+1}\rightarrow\mathbb{C}[/tex]​
    with one triple of coordinates for every particle and a time coordinate (usually) added? Correct me if I am wrong.

    Harald.
     
  8. Jun 15, 2008 #7

    vanesch

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Yes, that's correct for N scalar point particles in an Euclidean space.

    But one can think of many other quantum systems, such as spin systems, or fields, or strings, or I don't know what.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Eigenstates, Eigenblahs, and Eigenpoofs? HELP O.O
  1. Proving an eigenstate (Replies: 6)

  2. Common eigenstates (Replies: 7)

  3. Position eigenstates (Replies: 89)

Loading...