Einstein, Podolsky, and Rosen (1935)

In quantum mechanics -> if zz and xx Observable are incompatible, getting a Definite value for zz means losing the ability to know xx.

Einstein-Podolsky-Rosen (1935) argued that quantum mehcanics is incomplete as physics should obey principles of Local Realism

EPR postulated that let there be a pair of qubits in ∣Ψ−(AB)⟩\left\lvert \Psi_-^{(AB)} \right\rangle. Alice takes one and Bob takes one. They then separate light years from each other. Alice now measures her qubit in the z basis. Let’s say she gets 00.

∣Ψ−(AB)⟩=12(∣0(A)⟩⊗∣1(B)⟩−∣1(A)⟩⊗∣0(B)⟩)\left\lvert \Psi_-^{(AB)} \right\rangle=\frac{1}{\sqrt{2}}\left(\left\lvert {0}^{(A)} \right\rangle\otimes \left\lvert 1^{(B)} \right\rangle-\left\lvert 1^{(A)} \right\rangle\otimes \left\lvert {0}^{(B)} \right\rangle\right)

Multiplying by ⟨0(A)∣\left\langle {0}^{(A)} \right\rvert on the left, we get

⟨0(A)∣Ψ−(AB)⟩=(⟨0∣(A)⊗I(B))∣Ψ−(AB)⟩\left\langle {0}^{(A)} | \Psi_-^{(AB)} \right\rangle = \left(\left\langle 0 \right\rvert^{(A)} \otimes I^{(B)}\right) \left\lvert \Psi_-^{(AB)} \right\rangle =(⟨0∣(A)⊗I(B))⋅12(∣0(A)⟩⊗∣1(B)⟩−∣1(A)⟩⊗∣0(B)⟩)= \left(\left\langle 0 \right\rvert^{(A)} \otimes I^{(B)}\right) \cdot \tfrac{1}{\sqrt{2}}\left(\left\lvert {0}^{(A)} \right\rangle \otimes \left\lvert 1^{(B)} \right\rangle - \left\lvert 1^{(A)} \right\rangle \otimes \left\lvert {0}^{(B)} \right\rangle\right) =12[(⟨0∣(A)⊗I(B))(∣0(A)⟩⊗∣1(B)⟩)−(⟨0∣(A)⊗I(B))(∣1(A)⟩⊗∣0(B)⟩)]= \tfrac{1}{\sqrt{2}}\left[\left(\left\langle 0 \right\rvert^{(A)} \otimes I^{(B)}\right)\left(\left\lvert {0}^{(A)} \right\rangle \otimes \left\lvert 1^{(B)} \right\rangle\right) - \left(\left\langle 0 \right\rvert^{(A)} \otimes I^{(B)}\right)\left(\left\lvert 1^{(A)} \right\rangle \otimes \left\lvert {0}^{(B)} \right\rangle\right)\right] =12∣1(B)⟩=\frac{1}{\sqrt{2}}\left\lvert 1^{(B)} \right\rangle

That means, via. Born’s rule, if Alice observed a 00 outcome, Bob’s qubit is in the state ∣1⟩\left\lvert 1 \right\rangle with 100% probability.

EPR argues that it breaks local realism as local realism wants Bob’s qubit to have definite values of x and z observables.

Another example would be let there be two qualities of a particle: hardness {hard,soft}\{hard,soft\} and color {black,white}\{black, white\} if there is a particle with (z^+,x^+)(\hat{z}+, \hat{x}+) which just means it’s like (soft,white)(soft, white). This is not an eigenstate because in quantum mechanics, knowing z^=+\hat{z}=+ means losing the ability to know x^=−\hat{x}=- or x^=+\hat{x}=+. We’re just constructing a model that obeys EPR.

Let's take a Singlet We know that the EV of a singlet is -1:

E[σAZσBZ]=−1E[σAXσBX]=−1\mathbb{E}[\sigma_A^Z\sigma_B^Z]=-1\quad\mathbb{E}[\sigma_A^X\sigma_B^X]=-1

so if

σAZ=1⇒σBZ=−1\sigma_A^Z=1\Rightarrow\sigma_B^Z=-1 σAX=1⇒σBX=−1\sigma_A^X=1\Rightarrow \sigma_B^X=-1

Local realism assumes that before every measurement, every observable has a pre-existing definite value determined by some hidden variable λ\lambda. So in any given run, A1,A2,B1,B2A_1,A_2,B_1,B_2 all have predetermined definite values.

λAB (forced)
0(Z₊, X₊)(Z₋, X₋)
1(Z₊, X₋)(Z₋, X₊)
2(Z₋, X₊)(Z₊, X₋)
3(Z₋, X₋)(Z₊, X₊)
Each particle of a singlet is maximally mixed so every λ\lambda has a 25% change of occurring.

This complies with quantum mechanics but does this actually work for all measurements?

No! Bell's Theorem disproves EPR