A Quantum System can’t always be expressed as a tensor product of it’s constituents. Such systems are said to be in an entangled state (otherwise seperable state).

INTRODUCTION

Consider a two-particle system $\psi$ consisting of two particles $x$ & $y$, where each particle can be found in one of two possible states, as depicted by their state representation:

$\ket{x} = c_0\ket{x_0} + c_1\ket{x_1}$

$\ket{y} = c’_0\ket{y_0} + c’_1\ket{y_1}$

Further assume $\psi$ has the state:

$\ket{\psi} = \ket{x_0}\ket{y_0} + \ket{x_1}\ket{y_1}$

This, can then be re-written as

$\ket{\psi} = 1*\ket{x_0}\ket{y_0} + 0*\ket{x_0}\ket{y_1} + 0*\ket{x_1}\ket{y_0} + 1*\ket{x_1}\ket{y_1}$

Let’s now look at some results for the example at hand:

• $\ket{\psi}$ cannot be rewritten as a tensor product of its constituents.
• The individual states of the two particles $x$ & $y$ are intimately related to one another, or entangled.

EPR (Einstein Podolsky & Rosen)

Realism : According to EPR, Properties of physical systems have definite values (objective reality) whether you observe the system or not (sharply defined value pre-measurement).

Locality : Measurement of a paticle A in no way disturbs the state of a spatially separated particle B.

• Contradiction to the core value held by EPR:

  Consider, a qubit $\psi$ in state:

  **$\ket{\psi} = \ket{0} = \large \frac{\ket{+} + \ket{-}}{\sqrt{2}}$

  Post measurement wrt to operator $X$, the system does assume a definite state (either $\ket{+}$ or $\ket{-}$ with equal probability), however pre-measurement this is not the case. This is in direct contradiction to the values held by EPR (stated above).

• EPR demonstrated that quantum mechanics predicts that if two particles interact and then separate, measurement of one of the particles will determine the values that the properties of the other particles must assume. (This is the case, even if the particles are spatially separated and non-interacting at the time of measurement).

MEASUREMENTS OF POSITION & MOMENTUM

• For a given particle, position $x$ and momentum $p$ do not commute, i.e.:

  $[x, p] = i\hbar$

  • Proof : Two operators have simultaneous eigenstates (eigenvectors), if they commute.
  • Thus, using the above derived result, we can say that $x$ & $p$ do not have simultaneous eigenstates.
  • Hence, we concluded that if a definite value is assumed for one of the variables, then the other variable cannot be in a definite eigenstate.

• BASIC SCENARIO proposed by EPR:

• CONTRADICTION with Quantum Theory