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Spin singlets and the Bell states

  1. May 13, 2013 #1
    Given two spin-1/2 particles, the overall spin of the pair decomposes into a spin singlet and a spin triplet. Using the Clebsch-Gordon series and referring to the z-axis, we find the spin singlet is:

    ##|\Psi^- \rangle = \frac{1}{\sqrt{2}}(|\uparrow_z \downarrow_z \rangle - |\downarrow_z \uparrow_z \rangle)##

    The overall state is one of the Bell states and is given the standard label. Being a spin singlet is the same thing as being rotationally invariant. However, consider another of the Bell states:

    ##|\Phi^+ \rangle = \frac{1}{\sqrt{2}}(|\uparrow_z \uparrow_z \rangle + |\downarrow_z \downarrow_z \rangle)##

    This state has an interesting property; it may also be expressed as:

    ##|\Phi^+ \rangle = \frac{1}{\sqrt{2}}(|\uparrow_x \uparrow_x \rangle + |\downarrow_x \downarrow_x \rangle)##

    or, indeed,

    ##|\Phi^+ \rangle = \frac{1}{\sqrt{2}}(|\uparrow_{\hat{n}} \uparrow_{\hat{n}} \rangle + |\downarrow_{\hat{n}} \downarrow_{\hat{n}} \rangle)##

    where ##{\hat{n}}## is an arbitrary unit vector. But isn't this property just a mathematical expression of rotational invariance? So how does that square with the fact that it is ##|\Psi^- \rangle## (not ##|\Phi^+ \rangle##) that is the spin singlet for a bipartite system of two spin-1/2 particles?
     
  2. jcsd
  3. May 15, 2013 #2

    Avodyne

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    It's ##|\Psi^- \rangle## that has this property, not ##|\Phi^+ \rangle##. Work it out for the x case and you'll see.
     
  4. May 15, 2013 #3
    I have worked it out for the x case, and the result holds:
    ##| \uparrow_x \rangle = \frac{1}{\sqrt{2}}(| \uparrow_z \rangle + | \downarrow_z \rangle)##
    ##| \downarrow_x \rangle = \frac{1}{\sqrt{2}}(| \uparrow_z \rangle - | \downarrow_z \rangle)##

    Hence:
    ##\frac{1}{\sqrt{2}}(| \uparrow_x \uparrow_x \rangle + | \downarrow_x \downarrow_x \rangle ##
    ##=\frac{1}{2\sqrt{2}}((| \uparrow_z \rangle + | \downarrow_z \rangle)(| \uparrow_z \rangle + | \downarrow_z \rangle) + (| \uparrow_z \rangle - | \downarrow_z \rangle)(| \uparrow_z \rangle - | \downarrow_z \rangle))##
    ##=\frac{1}{\sqrt{2}}(| \uparrow_z \uparrow_z \rangle + | \downarrow_z \downarrow_z \rangle ##

    The general statement is proved (and used extensively by) Preskill in his notes (p.31)
     
  5. May 15, 2013 #4
    Moreover, this property does not hold for ##|\Psi^- \rangle##, it picks up an overall minus sign. This is irrelevant quantum mechanically; however, it is still odd, given it is supposed to be rotationally invariant. I'm more put off by the behaviour of ##|\Phi^+ \rangle## though, it seems like it shouldn't do this. But, as Preskill shows, it does.
     
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