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Trouble understanding the wave function

  1. Oct 27, 2015 #1
    I have been following Leonard Susskind's 'Theoretical Minimum' lecture series on quantum mechanics he made in winter 2012, and have, at least up to lecture 7/8, understood what he is doing - he has primarily been looking at systems of single spin 1/2 particles and pairs of them, examining phenomena such as entanglement along the way. I am happy with what a quantum state is, and how it can be represented as a vector.

    However, Susskind started to introduce the wave function to his lectures, and I suddenly became incredibly confused. He says the following:

    "The wave function of a system, in a particular basis, is the projection of the state onto the basis vectors"

    His equation for the wave function, ψ(a,b,c...), is: ψ(a,b,c...) = <a,b,c...|Ψ>, where |Ψ> is the state of the system.

    All I'm really asking for are examples of the wave functions for single and double spin 1/2 systems, simply so that I can understand what he is talking about.

    Thanks in advance.
     
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  3. Oct 27, 2015 #2

    Geofleur

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    It may help to first think in terms of the most familiar kinds of vectors, arrows in the plane. Suppose we have a vector ##\mathbf{v}=v_x\mathbf{i}+v_y\mathbf{j}##. Notice that ## \mathbf{i}\cdot \mathbf{v} = v_x \mathbf{i}\cdot \mathbf{i} + v_y \mathbf{i}\cdot\mathbf{j} = v_x ##; similarly, ## \mathbf{j} \cdot \mathbf{v} = v_y ##. That means we can equally well write ## \mathbf{v} = (\mathbf{i}\cdot \mathbf{v})\mathbf{i} + (\mathbf{j}\cdot\mathbf{v})\mathbf{j}##.

    Now let's represent the vectors the way Dirac would and write ## \mathbf{v} = v_x\mathbf{i} + v_y\mathbf{j} = | v\rangle = v_x|i\rangle + v_y|j\rangle ##. Instead of writing ## \mathbf{i} \cdot \mathbf{v} ##, Dirac would write ## \langle i | v \rangle ##, so that the last equation from above now reads ## | v \rangle = \langle i | v \rangle |v \rangle + \langle j | v \rangle |v \rangle##. The numbers ## v_x = \langle i | v \rangle ## and ## v_y = \langle j | v \rangle ## are just the components of the vector ## \mathbf{v}=|v\rangle## projected onto the basis ## \mathbf{i} = | i \rangle, \mathbf{j} = |j\rangle##.

    We can do the same thing with a quantum state. If ## | \uparrow \rangle ## represents a spin up state, and ## | \downarrow \rangle ## represents a spin down state, we can write an arbitrary state as ## |\Psi\rangle = \langle\uparrow | \Psi \rangle |\uparrow\rangle + \langle\downarrow | \Psi \rangle | \downarrow \rangle ##. Then the complex numbers ## \langle \uparrow | \Psi \rangle ## and ## \langle \downarrow | \Psi \rangle ## represent ## | \Psi \rangle ## in the basis ## | \uparrow \rangle ##, ## | \downarrow \rangle ##.
     
  4. Oct 27, 2015 #3
    Thanks, I understand a bit better now, but what would the wave function be? I guess if we write the state as |Ψ> = a|↑> + b|↓>, then ψ(1)=a and ψ(-1)=b, but I still don't really understand what the general wave function is, and certainly have no clue about the two spin system.
     
  5. Oct 27, 2015 #4

    Geofleur

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    We could represent the basis vectors with column matrices. So, for example, we could have ## | \uparrow \rangle = [ 1 \ 0 ]^T ## and ## | \downarrow \rangle = [ 0 \ 1 ]^T ##.
     
  6. Oct 27, 2015 #5

    DrClaude

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    That is one of the reasons why the Dirac notation is really useful, because you can't express spin as a regular function. I've seen notations like ##\alpha## for spin up and ##\beta## for spin down, such that your example would be
    $$
    \Psi = a \alpha + b \beta
    $$
    with the idea that you can also include a spatial part, such as
    $$
    \Psi = \psi(\mathbf{r}) \alpha
    $$
    for a system with the spatial wave function ##\psi(\mathbf{r})## in the spin state ##\alpha##.

    Another example I've seen is ##\chi^+## and ##\chi^-## for spin up and spin down. But in all cases, I feel that these notations are a bit clumsy and much prefer the Dirac notation.
     
  7. Oct 27, 2015 #6
    I understand this too, and do usually use column vectors to represent states in my own working, but what I really want to know is what the wave function for a spin in the state \begin{bmatrix} \alpha \\ \beta \end{bmatrix} is, as Susskind seems to be implying that there is such a thing, and I simply don't know what it is.
     
  8. Oct 27, 2015 #7

    Mentz114

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    Is it a spinor ##\Psi =\begin{bmatrix} \psi_- \\\psi_+ \end{bmatrix}##

    Quantum spin inhabits a 2-dimensional Hilbert space.
     
  9. Oct 27, 2015 #8

    Geofleur

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    Yes, and in relativistic quantum, the spinor has four components - two of them represent an ordinary particle and the other two represent the corresponding anti-particle with its spin states.
     
  10. Oct 27, 2015 #9
    I'm still not sure I really understand - is that spinor the wave function? If so, how is it constructed in the way Susskind says it can be?
     
  11. Oct 27, 2015 #10

    Mentz114

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    This paper might help.

    Michel Gondran, Alexandre Gondran
    A Complete analysis of the Stern-Gerlach experiment using Pauli spinors
    http://arxiv.org/pdf/quant-ph/0511276.pdf
     
    Last edited: Oct 27, 2015
  12. Oct 27, 2015 #11
    I think I'm becoming even more confused - I don't understand why spinors suddenly have anything to do with the problem I have. I simply ask how a wavefunction for a general state of one (or two) spin 1/2 particles can be constructed using the method Susskind explained.
     
  13. Oct 27, 2015 #12
    Tomdodd, I think I'm about at your level - I have Susskind's book but I haven't looked at the online lectures. What I'm about to say will probably make the real physicists and mathematicians here do backflips at their computers, but the way I understand it is, when I read 'basis' I think of an arbitrary Cartesian coordinate system, I guess technically with unit vectors along each axis. When I read 'projection onto the basis vectors' I think of the Cartesian coordinates of a vector in that particular coordinate system. Since spin states are described in two dimensions, the wave function as Susskind describes it has two coordinates, or projections, whatever. For an arbitrary state in the 'up/down' basis you would have an x up-coordinate/projection and a y down-coordinate/projection, where x squared + y squared = 1. I think. Let the flames begin.
     
  14. Oct 27, 2015 #13
    I am nevertheless unsure of what the wave function actually is for a single spin in an arbitrary state, [α,β], using the basis [1,0],[0,1], if such a thing exists - I assume one does as Susskind seems to imply there is one, but he never shows it.

    What my limited knowledge tells me is that ψ(ћ/2) = α and ψ(-ћ/2) = β, and this seems to work because ψ*(ћ/2)ψ(ћ/2) + ψ*(-ћ/2)ψ(-ћ/2) = α*α + β*β = 1.
     
    Last edited: Oct 27, 2015
  15. Oct 27, 2015 #14

    stevendaryl

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    Actually, matrices are easier to understand in Susskind's way than the sort of wave functions that you see in talking about the nonrelativistic Schrodinger's equation. The states are of the form [itex]|\psi \rangle = \left( \begin{array}\\ \alpha \\ \beta \end{array} \right)[/itex]. Then there are corresponding projection operators [itex]\langle +\frac{1}{2}| = \left( \begin{array}\\ 1 & 0 \end{array} \right)[/itex] and [itex]\langle -\frac{1}{2}| = \left( \begin{array}\\ 0 & 1 \end{array} \right)[/itex]. So:

    [itex]\langle +\frac{1}{2} | \psi \rangle = \left( \begin{array}\\ 1 & 0 \end{array} \right) \left( \begin{array} \\ \alpha \\ \beta \end{array} \right) = \alpha [/itex]

    [itex]\langle -\frac{1}{2} | \psi \rangle = \left( \begin{array}\\ 0 & 1 \end{array} \right) \left( \begin{array} \\ \alpha \\ \beta \end{array} \right) = \beta [/itex]

    You can combine those into a "function" [itex]\psi(s)[/itex] where [itex]s[/itex] ranges over two possible values: [itex]\pm \frac{1}{2}[/itex], where

    [itex]\psi(+\frac{1}{2}) = \langle +\frac{1}{2} | \psi \rangle = \alpha[/itex]
    [itex]\psi(-\frac{1}{2}) = \langle -\frac{1}{2} | \psi \rangle = \beta[/itex]

    In the case of the wave function for Schrodinger's equation, the variable [itex]x[/itex] ranges over infinitely many possible values, while in the spinor case, the variable [itex]s[/itex] only ranges over two possible values. But it's the same principle: For the Schrodinger equation's wave function, [itex]\langle a | \psi \rangle[/itex] is the value of [itex]\psi(x)[/itex] when [itex]x=a[/itex]. Similarly, for a spinor, [itex]\langle +\frac{1}{2} | \psi \rangle[/itex] = the value of [itex]\psi(s)[/itex] when [itex]s = +\frac{1}{2}[/itex].
     
  16. Oct 27, 2015 #15
    Right, I understand that (I'm guessing that you have set ћ=1). Since that is a simple system, I think I understand what the wave function is. However, where I originally got into a tangle (before opening this thread) was the wave function describing a system of two spin 1/2 particles. I'm not even sure which basis/bases to use.
     
  17. Oct 27, 2015 #16

    stevendaryl

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    Well, a state for a two-particle system is a kind of "product" of states for the one-particle systems. So if you have two spin-1/2 particles, [itex]A[/itex] and [itex]B[/itex], then a basis for [itex]A[/itex] is [itex]|s_A\rangle= | \pm \frac{1}{2}\rangle[/itex] and a basis for [itex]B[/itex] is [itex]|s_B \rangle = | \pm \frac{1}{2}\rangle[/itex]. So a basis for the two-component system is all possible values for [itex]|s_A\rangle \otimes |s_B\rangle[/itex]. So there are 4 elements in the basis:

    [itex]|+\frac{1}{2}\rangle \otimes |+\frac{1}{2}\rangle[/itex] (which corresponds to [itex]s_A = +\frac{1}{2}, s_B = +\frac{1}{2}[/itex])

    [itex]|+\frac{1}{2}\rangle \otimes |-\frac{1}{2}\rangle[/itex] (which corresponds to [itex]s_A = +\frac{1}{2}, s_B = -\frac{1}{2}[/itex])

    [itex]|-\frac{1}{2}\rangle \otimes |+\frac{1}{2}\rangle[/itex] (which corresponds to [itex]s_A = -\frac{1}{2}, s_B = +\frac{1}{2}[/itex])

    [itex]|-\frac{1}{2}\rangle \otimes |-\frac{1}{2}\rangle[/itex] (which corresponds to [itex]s_A = -\frac{1}{2}, s_B = -\frac{1}{2}[/itex])

    Usually, people leave out the [itex]\otimes[/itex] and just write:

    [itex]|\pm \frac{1}{2}, \pm \frac{1}{2}\rangle[/itex]

    A general two-particle spinor is a superposition of these four possibilities:

    [itex]|\psi \rangle = \alpha |+\frac{1}{2}, +\frac{1}{2} \rangle + \beta |+\frac{1}{2}, -\frac{1}{2} \rangle + \gamma |-\frac{1}{2}, +\frac{1}{2} \rangle + \delta |-\frac{1}{2}, -\frac{1}{2} \rangle[/itex]

    In Susskind's language,

    [itex]\alpha = \langle +\frac{1}{2}, +\frac{1}{2} | \psi \rangle[/itex]
    [itex]\beta= \langle +\frac{1}{2}, -\frac{1}{2} | \psi \rangle[/itex]
    [itex]\gamma = \langle -\frac{1}{2}, +\frac{1}{2} | \psi \rangle[/itex]
    [itex]\delta = \langle -\frac{1}{2}, -\frac{1}{2} | \psi \rangle[/itex]
     
  18. Oct 27, 2015 #17
    So what would the wave function corresponding to an entangled state, such as the singlet state, be?
     
  19. Oct 27, 2015 #18

    stevendaryl

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    A state corresponds to specific values for [itex]\alpha, \beta, \gamma[/itex] and [itex]\delta[/itex]. So the singlet state would be the case where:

    [itex]\alpha = 0, \beta = +\frac{1}{\sqrt{2}}, \gamma = -\frac{1}{\sqrt{2}}, \delta = 0[/itex]

    Which is the state: [itex]|\psi\rangle = \frac{1}{\sqrt{2}} (|+\frac{1}{2}, -\frac{1}{2}\rangle - |-\frac{1}{2}, +\frac{1}{2}\rangle)[/itex]
     
  20. Oct 27, 2015 #19
    So if the state of the two spin 1/2 system is the singlet state, [itex]\frac{1}{\sqrt{2}} (|\frac{ћ}{2}, -\frac{ћ}{2}\rangle - |-\frac{ћ}{2}, \frac{ћ}{2}\rangle)[/itex], would the following be true:

    [itex]\psi(\frac{ћ}{2},\frac{ћ}{2}) = 0[/itex]
    [itex]\psi(\frac{ћ}{2},-\frac{ћ}{2}) = \frac{1}{\sqrt{2}}[/itex]
    [itex]\psi(-\frac{ћ}{2},\frac{ћ}{2}) = -\frac{1}{\sqrt{2}}[/itex]
    [itex]\psi(-\frac{ћ}{2},-\frac{ћ}{2}) = 0[/itex] ?
     
  21. Oct 27, 2015 #20

    stevendaryl

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    Yes, that's correct, but people don't usually use the function notation [itex]\psi(\frac{ћ}{2},\frac{ћ}{2})[/itex] for discrete values.
     
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