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  what is the difference between product states and entangled states?

+ 0 like - 0 dislike

now , i know that product states can be factorisable into two separate superpositions.

e:g (a l0> + b l1>) * (c l0> + d l1>) = ac l00> + ad l01> + bc l10> +  bd l11>    how does this differ from entangled states which also has 4 coefficients (e:g) a l00> + b l01> + c l10> + d l11>.

i know that entangled states aren't factorizable but how do they differ from product states if they end up with the same number of coefficients. like why does entangled states give us  2^n computational states while superrpositions without entanglement give us 2n although as written above its the same number of coefficents?

if we add a third qubit entangled states will have 8 coefficients and product states 6, though both will give all possible combinations of the three qubit system (e:g 000, 001, 011, .....) when we multiply out the product states made of three separate superpositions. how does that make any difference?

asked May 17, 2020 in Theoretical Physics by yousef [ no revision ]
recategorized May 17, 2020 by Dilaton

1 Answer

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Take $(a\left|0\right>+b\left|1\right>)(c\left|0\right>+d\left|1\right>)$.

If $a\neq0$, $c\neq 0$, $ac(\left|0\right>+\frac{b}{a}\left|1\right>)(\left|0\right>+\frac{d}{c}\left|1\right>)$, so you have 3 degrees of freedom: $ac$, $\frac{b}{a}$ and $\frac{d}{c}$.

Take $a\left|00\right>+b\left|10\right>+c\left|01\right>+d\left|11\right>$.

If $a\neq0$, $c\neq 0$, $a\left|00\right>+b\left|10\right>+c\left|01\right>+d\left|11\right>$ and there are 4 degrees of freedom.

answered May 19, 2020 by Iliod (35 points) [ revision history ]
edited May 19, 2020 by Iliod

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