Monday, 4 July 2016

quantum mechanics - Set of states |phinrangle in the density operator rho=sumlimits_n p_n|phi_nranglelanglephi_n|


The set of quantum states \{|\phi_n\rangle\} in the definition of the density operator \rho=\sum\limits_n p_n|\phi_n\rangle\langle\phi_n| need not be orthonormal, and need not form a basis. But unfortunately, in the examples that I have seen so far, the states \{|\phi_n\rangle\} were both orthonormal and forms a basis.


Example 1 In the Stern-Gerlach (SG) set-up, the state of the silver atoms coming out of the oven and before passing through the magnetic field, is imperfectly known because S_z remained unmeasured. Therefore, on the ignorance ground, such an ensemble will be represented by \rho=\frac{1}{2}(|{\uparrow}\rangle\langle{\uparrow}|+|{\downarrow}\rangle \langle{\downarrow}|).\tag{1} Note that, in this case, the states |{\uparrow}\rangle and |{\downarrow}\rangle are orthonormal and forms the S_z-basis.


Example 2 Consider an unpolarized light moving in the z-direction so that its polarization must be in the xy-plane. Since we do not know the state vector, it is described by the density operator \rho=\frac{1}{2}(|x\rangle\langle x|+|y\rangle\langle y|)\tag{2} where |x\rangle and |y\rangle describe plane polarized states along the x and y-axes respectively.





Question Can someone suggest an example of a mixed ensemble where the states \{|\phi_n\rangle\} need not be orthonormal and need not form a basis? I'm not looking for the trivial example where the desity operator describes a pure state.




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