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What is the difference between states

$$ \frac1{\sqrt{2}} |11\rangle+\frac1{\sqrt{2}} |00\rangle $$

and

$$ \frac1{\sqrt{2}} |11\rangle- \frac1{\sqrt{2}} |00\rangle~? $$

They will all eventually mean probabilities. So what difference does it make?

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This question gets to the heart of what makes quantum mechanical amplitudes different from classical probabilities. It is true that if you make measurements in the basis of states $\{\lvert 00 \rangle ,\lvert 11 \rangle\}$ then the two states have the same measurement statistics, and so cannot be distinguished. The interesting thing is that it is possible to measure in other bases. In particular, there exists a measurement in the basis $\{(\lvert 00 \rangle\ + \lvert 11 \rangle)/\sqrt{2},(\lvert 00 \rangle\ - \lvert 11 \rangle)/\sqrt{2}\}$ which can perfectly distinguish the two states.

Note that this issue has nothing to do with entangled or two-particle states in particular. Exactly the same could be said of the states $$\lvert \pm\rangle = \frac{1}{\sqrt{2}} ( \lvert 0\rangle \pm \lvert 1 \rangle ).$$ These states are indistinguishable when measured in the $\{ \lvert 0\rangle,\lvert 1 \rangle\}$ basis, but perfectly distinguishable in the $\{ \lvert +\rangle,\lvert - \rangle\}$ basis.

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When you define a base of vectors, in your case $\{ |1,1\rangle, |1,0\rangle, |0,1\rangle, |0,0\rangle \} $ you assign to each vector some phase, e.g. the vector $|1,1\rangle$ has a certain phase that you don't mention explicitly outside the bra-kets. Now, in your calculi you may have superpositions of these vectors of the form

$|\psi\rangle = ae^{i\alpha} |1,1\rangle + be^{i\beta}|1,0\rangle + ce^{i\gamma}|0,1\rangle + de^{i\delta}|0,0\rangle,$

with $a, b, c, d$ positive numbers, i.e. besides the intrinsic phases of the four vectors, they appear in the superposition with complex amplitudes. Thus the amplitude of $|0,0\rangle$ has a difference of phase in comparison with the amplitude of $|1,1\rangle$ and this difference is $\delta - \alpha$.

Returning to your functions, in the first superposition the vectors $|1,1\rangle$ and $|0,0\rangle$ are in phase, while in the 2nd superposition they are in anti-phase, there is a difference $\delta - \alpha = \pi$ between them, i.e. you can write

$ \frac {1}{\sqrt{2}}e^{i \cdot 0} |1,1\rangle + \frac {1}{\sqrt{2}}e^{i\pi}|0,0\rangle$ .

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The difference between those two states is in their phase $|\psi(\alpha)\rangle=\frac{1}{\sqrt{2}}(|11\rangle+\exp[i\alpha]|00\rangle)$ for $\alpha=2\pi$ you have the first one and for $\alpha=\pi$ you have the second one with the minus sign. Phsycally the difference lies in the interference pattern you might see once you compute the probability amplitudes, meaning $p_{\alpha}=\langle\psi(\alpha)|\psi(\alpha)\rangle=1$ if we define $p_{\alpha\beta}=\langle\psi(\alpha)|\psi(\beta)\rangle=\frac{1}{2}(1+\exp[i(\beta-\alpha)])$ you can see now that by varying $\alpha$ and $\beta$ you can adjust the degree of orthogonality of the two state vectors, for example if $\beta-\alpha=\pi$ the two states are orthogonal and $p_{\alpha\beta}=0$ in such case they make a base you can use for performing measures! Instead if they are not completely orthogonal there is a non zero statistical correlation between the two, meaning there is a non zero probability for one to interfere with the outcome of a measure of the other in an interferometric experiment: for istance take the two states $\psi(\alpha)$ and $\psi(\beta)$ and make them go through two different branches of an interferometer (a mach-zender maybe) than perform a measure at the end of it after the two states have interfered, what you will find in the computational basis $\{|00\rangle,|11\rangle\}$ (which you can think as the polarization $\{HH,VV\}$) is that the two states $\psi(\alpha)$ $\psi(\beta)$ by having non zero $p_{\alpha\beta}$ will affect the polarizations count of each other giving rise to interference patterns in such counts!

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Mostly likely you would see $$|01\rangle+|10\rangle$$ vs. $$|01\rangle-|10\rangle$$ since they have same Eigenvalue for energy and hard to distinguished unless there are another measure to tell the difference. $|00\rangle$ and $|11\rangle$ are very different in this sense.

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