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I'm trying to learn about the Bell state $\frac{1}{\sqrt{2}}|00\rangle+\frac{1}{\sqrt{2}}|11\rangle$. Question 10.1 in Algorithms asks us to show that this cannot be decomposed into the tensor product of two single qubits' states.
It seems to me however that this can be decomposed while still obeying the basic rules. Wolfram Alpha lists some solutions. What am I doing wrong?
All you need to show is that there do not exist any $a,b,c$ or $d$ satisfying the conditions
$\frac{1}{\sqrt{2}}(|00\rangle)+|11\rangle)=(a|0\rangle+b|1\rangle)\otimes(c|0\rangle+d|1\rangle)$
and $|a|^2+|b|^2=|c|^2+|d|^2=1$.
You should be able to do this quite easily!
This property of the tensor product has really nothing to do with its interpretation in quantum mechanics or your specific example. Given two vector spaces $U,V$ the tensorproduct $\phi: U \times V \to U \otimes V$ is characterized by the property (up to isomorphism), that for any bilinear map $b \colon U \times V \to W$ into some third vectorspace $W$, there exists a unique linear map $l \colon U \otimes V \to W$, such that $b = l\circ\phi$. Because the two projections from the direct product $U\times V \to U$ and $U \times V \to V$ are not bilinear, they do not yield a map $U\otimes V \to U$ or $U \otimes V \to V$.
In quantum mechanical parlance this is called entanglement. As you probably know in this case $U$ and $V$ are spaces of state and $U \otimes V$ is the space of state for the combined system. Notice that just because the projections do not yield maps $U \otimes V \to V$ this does not mean there are no such maps. If for instance $U = H_e$ is the state space of a relativistic electron and $V = H_\gamma$ the state space of a photon, there is a map $H_e \otimes H_\gamma \to H_e$ that describes the absorption of a photon by an electron. In this case one usually draws a feynman diagram with a wiggly line for the photon meeting the electron line.
