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Homework Help: Spin-1/2 systems, Spinor representation

  1. Nov 29, 2007 #1
    I'm totally confused on the relationship between kets written

    [tex]| \uparrow \rangle, | \downarrow \rangle[/tex]


    [tex]| \uparrow \uparrow \rangle, | \uparrow \downarrow \rangle | \downarrow \uparrow \rangle, | \downarrow \downarrow \rangle[/tex]

    (Problem) I have a system of two spin-1/2 particles, prepared in the singlet state, and I'd like to find the probability of measuring the first one and finding it to be spin-up. This being a homework problem from last week, I have the provided solution, but I don't understand it.

    (Attempt at Solution) I know it is 50%, simply because no other measurements have been made of the system, and there's an equal chance of it being up or down, given that the singlet state can be described as:

    [tex] |s=0,m=0\rangle = \frac{1}{\sqrt{2}} (| \uparrow \downarrow \rangle - | \downarrow \uparrow \rangle) [/tex]

    ...but I'd like to illustrate it using bra-ket notation. The solution says to project [tex] | \uparrow \rangle \langle \uparrow | [/tex] onto the possible eigenstates of S_z for the first particle

    [tex] (| \uparrow \rangle \langle \uparrow | + | \downarrow \rangle \langle \downarrow |)[/tex],

    to get...

    [tex]| \uparrow \uparrow \rangle \langle \uparrow \uparrow | + | \uparrow \downarrow \rangle \langle \uparrow \downarrow |[/tex]

    First off, "projecting" here means matrix multiplication, right? (or direct product?) Second, how do we move from kets representing spin for one particle, to kets representing spin for two particles?

    All help in understanding this is greatly appreciated!
    Last edited: Nov 29, 2007
  2. jcsd
  3. Nov 29, 2007 #2


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    Just apply the general rule: Find the operator P which projects on the eigenspace of S_z corresponding to the eigenvalue +1. (Actually the operator is [itex]S_z^{(1)}\otimes I^{(2)}[/itex]. We're leaving the second particle alone). Then, if your state is [itex]|\psi \rangle[/itex], the probability of finding +1 is then [itex]\langle \psi|P^{(1)}_{+1}| \psi \rangle[/itex].
    Going back to the postulates usually gets you where you want to go if you lost your way.

    You project a projection operator [tex] | \uparrow \rangle \langle \uparrow | [/tex]?
    Onto [tex] (| \uparrow \rangle \langle \uparrow | + | \downarrow \rangle \langle \downarrow |)[/tex]? This last expression is just the identity.

    What I'm pretty sure is meant is that you form the projection operator I mentioned earlier. The projector [itex]| \uparrow \rangle \langle \uparrow |[/itex] lives in the space of the first particle. The projector [tex] (| \uparrow \rangle \langle \uparrow | + | \downarrow \rangle \langle \downarrow |)=I^{(2)}[/tex] lives in the space of the second particle. You form the tensor product of those two, minding the order, since the labels for particles 1 and 2 are the same in the notation used:
    [tex] | \uparrow \rangle \langle \uparrow |\otimes (| \uparrow \rangle \langle \uparrow | + | \downarrow \rangle \langle \downarrow |) = | \uparrow \uparrow \rangle \langle \uparrow \uparrow | + | \uparrow \downarrow \rangle \langle \uparrow \downarrow |[/tex]
    This is the required projection. In general, if you measure [tex]A \otimes B[/tex], the probability of getting [tex](a,b)[/itex] is gotten from the projection operator [itex]P_a \otimes P_b[/itex], where the P's are ofcourse the projection upon the eigenspaces belonging to a and b.
    Looking at it from a more general viewpoint often elucidates the structure, especially if it's your first encounter with product spaces.
    You form the tensor product of the two state spaces. It consists of products of kets |1>|2>, where |1> is a ket for the first space and |2> for the second, and linear combinations thereof. I know some textbooks, like Griffiths, don't go into this very much, taking it as the obvious generalization to systems with higher degrees of freedom.

    PS: The following notations are all used frequently and mean the same thing:
    [tex]|v\rangle \otimes |w\rangle=|v\rangle|w\rangle=|v,w\rangle[/tex]
    Just keep in mind in which space each ket lives in if there is no specific label to distinguish them. In general the order doesn't matter, but without a label that is the only way to keep |v>|w> and |w>|v> apart.
    Last edited: Nov 29, 2007
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