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 $A$,

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


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

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?

  • $\begingroup$ LaTeX tip: \begin{pmatrix} is way easier than lugging around a bunch of \left( and \right)s. $\endgroup$ Dec 16, 2016 at 12:08
  • $\begingroup$ A small unimportant mistake : you forgot a $\frac{1}{\sqrt2}$ prefactor in front of your definition of $|±x\rangle$ $\endgroup$ Dec 16, 2016 at 17:04

1 Answer 1


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.

  • $\begingroup$ Why don't scalars (like the -1 in the singlet) distribute over both terms of the tensor product? $\endgroup$
    – dain
    Dec 17, 2016 at 2:25
  • $\begingroup$ (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? $\endgroup$
    – dain
    Dec 17, 2016 at 3:20

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