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4

Just to add, any system that can be in two states can serve as qubit. So any elementary particle having two or more values of spin can be a qubit. Moreover, another parameter of a particle can be used for representing a qubit, for example location (you can have one electron and you are looking where is is placed).


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A photon is a massless spin-1 particle. This means that a photon has exactly two spin states, just like an electron.


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If it is clear to you how to measure 'normal' (orthogonal) measurement then you can always implement any Povm as Take another system (ancilla) Act on both with a unitary transformation Perform a joint orthogonal measurment I will look up the paper proving this.


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I think the question will be much more clear if you specify some of the remaining basis vectors, for instance the $\vert{\uparrow 0}\rangle$. I recommend to write the state as follows. $$\vert{i}\rangle=\dfrac{1}{\sqrt{2}}(\vert{\uparrow 0}\rangle_A\vert{\downarrow 0}\rangle_B+\vert{\downarrow 0}\rangle_A\vert{\uparrow 0}\rangle_B)$$ Note that it lives in ...


4

The spatial part of the wave function is antisymmetric for the triplet and symmetric for the singlet. You can check this by considering the He atom. Its ground state is a singlet with two electrons in its 1s orbital. Clearly a doubly occupied 1s orbital is symmetric. The two fermion wave function must be antisymmetric under fermion exchange. Since the spin ...


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On a related note, let's say I have an arbitrary unitary matrix. How does one map that to a quantum gate? There is a several quantum gates you can use for construction other more complex gates. See list of gates on Wikipedia. Please also find other technique how to decompose arbitrary unitary gate in article Elementary gates for quantum computation. ...


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Measurement is usually just defined as a gate on it's own. The "well known gate" that measurement maps to is simply the measurement gate. Its one of the few places where some difficult, hard to model interactions with an environment are allowed in a quantum circuit, so we just separate that part out. For example, in this circuit You can clearly see two ...


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Your mistake is that you assume $e^{A\otimes B}=e^{A}\otimes e^{B}$. This is wrong - just as, in analogy, $e^{ab}=e^a\,e^b$ is wrong for numbers $a$ and $b$. (Indeed, if $A$ and $B$ are just numbers, tensor product boils down to the normal product.) A very useful fact is that for any $M$ with $M^2=I$, $$ e^{-iMt} = \cos(t)\,I - i\,\sin(t)\, M $$ which you ...


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The Hamiltonian can be written as $H = Z_A\otimes X_B = (|0_A\rangle\langle0_A|-|1_A\rangle\langle1_A|)\otimes X_B$, where we have expanded $Z_A = |0_A\rangle\langle 0_A|-|1_A\rangle\langle 1_A|$ in its eigenbasis. the evolution operator can be written as $$U_{AB} = e^{-iHt} = e^{-i(|0_A\rangle\langle 0_A|-|1_A\rangle\langle 1_A|)\otimes X_B t}$$ $$ = \...


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