Spin singlets and the Bell states

[VantagePoint72]

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?

 

[Avodyne]

It's ##|\Psi^- \rangle## that has this property, not ##|\Phi^+ \rangle##. Work it out for the x case and you'll see.

 

[VantagePoint72]

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 http://www.theory.caltech.edu/people/preskill/ph229/notes/chap2.pdf (p.31)

 

[VantagePoint72]

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.