# Spin singlets and the Bell states

1. May 13, 2013

### 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?

2. May 15, 2013

### Avodyne

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

3. May 15, 2013

### 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 his notes (p.31)

4. May 15, 2013

### 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.