# Bell's Theorem: Help with understanding final bit of wikipedia proof (tensor products)

I'm trying to understand this derivation of Bell's Theorem on Wikipedia.

I've never encountered tensor products before, but from my reading I gather that the first bit of the tensor product acts on the first particle, and the second bit acts on the second particle. I tried using the Pauli matrices to work through the problem explicitly but I can't seem to get the $\frac{1}{\sqrt{2}}$ result. This is what I did:

$$\left|+x\right\rangle = \left( \begin{matrix}1\\ 1 \end{matrix} \right),\ \left|-x\right\rangle = \left( \begin{matrix}1\\ -1 \end{matrix} \right)$$

\begin{align} |\psi\rangle & = \frac{1}{\sqrt{2}}\left( \left|+x\right\rangle \otimes\left|-x\right\rangle -\left|-x\right\rangle \otimes\left|+x\right\rangle \right) \\ & = \frac{1}{\sqrt{2}}\left( \left( \begin{matrix}1\\ 1 \end{matrix} \right) \otimes \left( \begin{matrix}1\\ -1 \end{matrix} \right) - \left( \begin{matrix}1\\ -1 \end{matrix} \right) \otimes \left( \begin{matrix}1\\ 1 \end{matrix} \right) \right) \\ & = \frac{1}{\sqrt{2}}\left( \left( \begin{matrix}1\\ 1 \end{matrix} \right) \otimes \left( \begin{matrix}1\\ -1 \end{matrix} \right) + \left( \begin{matrix}-1\\ 1 \end{matrix} \right) \otimes \left( \begin{matrix}-1\\ -1 \end{matrix} \right) \right) \end{align}

The expected value of operators $AB$ is

$$\langle AB\rangle = \langle \psi |AB| \psi\rangle$$

In Wikipedia's notation I'm considering the case $(a,b)$, with operators $A=A(a)=S_z\otimes I$ and $B=B(b)=-\frac{1}{\sqrt{2}}I \otimes (S_X+S_Z)$. The operator B isthen

$$B = -\frac{1}{\sqrt{2}}I \otimes (S_Z + S_X) = -\frac{1}{\sqrt{2}} \left( \left( \begin{matrix}1 & 0\\ 0 & 1 \end{matrix} \right)\otimes \left( \begin{matrix}1 & 1\\ 1 & -1 \end{matrix} \right) \right)$$

so when operated on $\psi$ I get $$B|\psi \rangle = -\frac{1}{2}\left[\left( \begin{matrix}1\\ 1 \end{matrix} \right) \otimes \left( \begin{matrix}0\\ 2 \end{matrix} \right) + \left( \begin{matrix}-1\\ 1 \end{matrix} \right) \otimes \left( \begin{matrix}-2\\ 0 \end{matrix} \right)\right]$$

As for $A$,

$$A = S_Z \otimes I = \left( \begin{matrix}1 & 0\\ 0 & -1 \end{matrix} \right) \otimes \left( \begin{matrix}1 & 0\\ 0 & 1 \end{matrix} \right)$$

so

$$AB|\psi \rangle = -\frac{1}{2}\left[ \left( \begin{matrix}1\\ -1 \end{matrix} \right) \otimes \left( \begin{matrix}0\\ 2 \end{matrix} \right) +\left( \begin{matrix}-1\\ -1 \end{matrix} \right) \otimes \left( \begin{matrix}-2\\ 0 \end{matrix} \right)\right].$$

Then to do the inner product I need to make the bra $\langle \psi|$ by flipping everything inside into row vectors:

$$\langle \psi| = \frac{1}{\sqrt{2}} \left[ (1\ \ \ 1)\ \otimes\ (1\ \ -1)\ + (-1\ \ \ 1)\ \otimes\ (-1\ \ -1) \right]$$

Finally, I get $$\langle \psi | AB | \psi \rangle = -\frac{1}{2\sqrt{2}}\left[ 0 \otimes -2 + 0 \otimes 2 \right] = ?$$

At this point I don't know where I've gone wrong. I don't know what the value in the square brackets is (do they just multiply like normal numbers now?), but I don't see any way for it to be $-2,$ which is what it needs to be in order to cancel properly and make the thing equal $\frac{1}{\sqrt2}$. Am I not doing the inner product properly here? Have I completely misunderstood how tensors work?

• LaTeX tip: \begin{pmatrix} is way easier than lugging around a bunch of \left( and \right)s. – Emilio Pisanty Dec 16 '16 at 12:08
• A small unimportant mistake : you forgot a $\frac{1}{\sqrt2}$ prefactor in front of your definition of $|±x\rangle$ – Frédéric Grosshans Dec 16 '16 at 17:04

You are unduly dropping some terms when you take the inner product between $⟨\psi|$ and $AB|\psi⟩$. You need to multiply everything against everything, because the inner product distributes over the sums. Schematically, you need to do \begin{align} \bigg(⟨a|\otimes⟨b|+⟨c|\otimes⟨d|\bigg) & · \bigg(|e⟩\otimes |f⟩+|g⟩\otimes|h⟩\bigg) \\ &= ⟨a|\otimes⟨b|·|e⟩\otimes |f⟩+⟨c|\otimes⟨d|·|g⟩\otimes|h⟩ \\ & \quad +⟨a|\otimes⟨b|·|g⟩\otimes|h⟩+⟨c|\otimes⟨d|·|e⟩\otimes |f⟩ \\ &= ⟨a|e⟩⟨b|f⟩+⟨c|g⟩⟨d|h⟩ \\ & \quad +⟨a|g⟩⟨b|h⟩+⟨c|e⟩⟨d|f⟩ . \end{align} You've done the $⟨a|e⟩⟨b|f⟩$ and $⟨c|g⟩⟨d|h⟩$ terms, but you're missing the $⟨a|g⟩⟨b|h⟩$ and $⟨c|e⟩⟨d|f⟩$ ones.

In your example, your calculation of the two sides of the inner product, $$AB|\psi \rangle = -\frac{1}{2}\left[ \begin{pmatrix}1\\-1\end{pmatrix} \otimes \begin{pmatrix}0\\2\end{pmatrix} +\begin{pmatrix}-1\\-1\end{pmatrix} \otimes \begin{pmatrix}-2\\0\end{pmatrix} \right]$$ and $$\langle \psi| = \frac{1}{\sqrt{2}} \left[ \begin{pmatrix}1&1\end{pmatrix} \otimes \begin{pmatrix}1&-1\end{pmatrix} +\begin{pmatrix}-1&1\end{pmatrix} \otimes \begin{pmatrix}-1&-1\end{pmatrix} \right]$$ is correct in its broad strokes. However, you overdid the signs when absorbing the singlet's substraction into both terms of the tensor product (instead of just one), and you missed a beat in the normalization of your basis states, which should read $$\left|+x\right\rangle =\frac{1}{\sqrt{2}} \begin{pmatrix}1\\1\end{pmatrix} ,\ \left|-x\right\rangle =\frac{1}{\sqrt{2}} \begin{pmatrix}1\\-1\end{pmatrix}.$$ This changes your two inner-product factors to $$AB|\psi \rangle = -\frac{1}{4}\left[ \begin{pmatrix}1\\-1\end{pmatrix} \otimes \begin{pmatrix}0\\2\end{pmatrix} +\begin{pmatrix}1\\1\end{pmatrix} \otimes \begin{pmatrix}-2\\0\end{pmatrix} \right]$$ and $$\langle \psi| = \frac{1}{2\sqrt{2}} \left[ \begin{pmatrix} 1&1\end{pmatrix} \otimes \begin{pmatrix}1&-1\end{pmatrix} +\begin{pmatrix}-1&1\end{pmatrix} \otimes \begin{pmatrix}1& 1\end{pmatrix} \right],$$ respectively, and introduces a global factor of $1/4$ with respect to your current result.

For the first inner product (i.e. the $⟨a|e⟩⟨b|f⟩$ term), then, you get \begin{align} \begin{pmatrix}1&1\end{pmatrix} \otimes \begin{pmatrix}1&-1\end{pmatrix} · \begin{pmatrix}1\\-1\end{pmatrix}\otimes \begin{pmatrix}0\\2\end{pmatrix} & = \begin{pmatrix}1&1\end{pmatrix} ·\begin{pmatrix}1\\-1\end{pmatrix} \times \begin{pmatrix}1&-1\end{pmatrix}·\otimes \begin{pmatrix}0\\2\end{pmatrix} \\ & = 0\times(-2) \\ & = 0. \end{align} Here the two single-particle inner products $⟨a|e⟩⟨b|f⟩$ just multiply as complex numbers - they are just complex numbers. Zero times whatever is always zero.

The other term in your calculation is also zero: \begin{align} \begin{pmatrix}-1&1\end{pmatrix} \otimes \begin{pmatrix}1&1\end{pmatrix} · \begin{pmatrix}1\\1\end{pmatrix}\otimes \begin{pmatrix}-2\\0\end{pmatrix} & = \begin{pmatrix}-1&1\end{pmatrix} ·\begin{pmatrix}1\\1\end{pmatrix} \times \begin{pmatrix}1&1\end{pmatrix}·\otimes \begin{pmatrix}-2\\0\end{pmatrix} \\ & = 0\times (-2) \\ & = 0. \end{align} However, there are also two other terms that contribute: \begin{align} \begin{pmatrix}1&1\end{pmatrix} \otimes \begin{pmatrix}1&-1\end{pmatrix} · \begin{pmatrix}1\\1\end{pmatrix}\otimes \begin{pmatrix}-2\\0\end{pmatrix} & = \begin{pmatrix}1&1\end{pmatrix} ·\begin{pmatrix}1\\1\end{pmatrix} \times \begin{pmatrix}1&-1\end{pmatrix}·\otimes \begin{pmatrix}-2\\0\end{pmatrix} \\ & = 2 \times (-2) \\ & = -4, \end{align} and \begin{align} \begin{pmatrix}-1&1\end{pmatrix} \otimes \begin{pmatrix}1&1\end{pmatrix} · \begin{pmatrix}1\\-1\end{pmatrix}\otimes \begin{pmatrix}0\\2\end{pmatrix} & = \begin{pmatrix}-1&1\end{pmatrix} ·\begin{pmatrix}1\\-1\end{pmatrix} \times \begin{pmatrix}1&1\end{pmatrix}·\otimes \begin{pmatrix}0\\2\end{pmatrix} \\ & = (-2) \times 2 \\ & = -4. \end{align}

Putting all of this together, then, we get $$⟨\psi|AB|\psi⟩ =\frac{-1}{8\sqrt{2}}(-4-4) =\frac{1}{\sqrt{2}},$$ as it needs to be.

Oh, and one final hint: these calculations are way easier if you do them on the canonical basis, $$\left|+x'\right\rangle =\begin{pmatrix}1\\0\end{pmatrix} ,\ \left|-x'\right\rangle =\begin{pmatrix}0\\1\end{pmatrix}.$$ using the more standard expression for the spin singlet state, $$|\psi' \rangle = \begin{pmatrix}1\\0\end{pmatrix} \otimes \begin{pmatrix}0\\1\end{pmatrix} -\begin{pmatrix}0\\1\end{pmatrix}\otimes \begin{pmatrix}1\\0\end{pmatrix} .$$ It will be a good exercise in the handling of tensor products to show that this $|\psi'⟩$ coincides with the $|\psi⟩$ defined above.

• Why don't scalars (like the -1 in the singlet) distribute over both terms of the tensor product? – dain Dec 17 '16 at 2:25
• (adding another comment because I can't edit) Also shouldn't it be $|+z'\rangle$ and $|-z'\rangle$ at the end, since those are the eigenvectors of the $S_Z$ operator? – dain Dec 17 '16 at 3:20