p212ln29 - Physics 212 Statistical mechanics II Fall 2006...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
Physics 212: Statistical mechanics II, Fall 2006 Lecture XXX Quantum information Here we give a quick overview of a few ideas from quantum information that have applications in statistical physics. An interesting property of quantum mechanics is that a pure state | Ψ AB i of a system partitioned into subsystems A and B does not always imply that A and B are in pure quantum states. If A and B are individually in pure quantum states, we say that | Ψ AB i is a product state: | Ψ AB i = | ψ A i ⊗ | ψ B i . (1) An example is the state of two spin-half degrees of freedom in which both spins A and B are up: | ↑ A B i ≡ | ↑ A i ⊗ | ↑ B i is a product state. However, many (in fact “most”) states are not product states: a familiar example is the singlet state of two spins: 1 2 ( | ↑ A B i - | ↓ A B i ) . (2) Note that the S z = 0 symmetric triplet state would also be entangled, even though it is part of the same angular momentum multiplet as the product (unentangled states) with S z = ± ¯ h . When the overall pure state of AB , | Ψ AB i , is not a product state, the description of a subsystem (say A ) must be as a density matrix rather than a pure state. The “reduced density matrix” for subsystem A is defined as (here | ψ i i are a basis for A , and | φ i i are a basis for B ): h ψ i | ρ A | ψ j i ≡ X k ( h ψ i | ⊗ h φ k | ) | Ψ AB ih Ψ AB | ( | ψ j i ⊗ | φ k i ) . (3) We now show that the same value of S is obtained whether we keep subsystem A and trace over B , or vice versa: S ( ρ A ) = S ( ρ B ) . (4) Proof: We will show a slightly stronger statement, that under the conditions assumed above, the two reduced density matrices have the same nonzero eigenvalues. Since the two in general have different dimensionalities, there may be more zero eigenvalues in one than in the other, but the entropy depends only on the nonzero eigenvalues and hence will be the same. This result follows from the Schmidt decomposition, that | Ψ AB i can be written, for some orthonormal bases for A and B , as | Ψ AB i = n X i =1 λ i | i A i| i B i . (5) Here λ i are real numbers with n X i =1 λ i 2 = 1 , (6) and the number n of nonzero values is known as the Schmidt number. The Schmidt decomposition implies that λ i are the eigenvalues both of ρ A and ρ B , as ρ A = X i λ i 2 | i A ih i A | , ρ B = X i λ i 2 | i B ih i B | . (7) 1
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
To prove the Schmidt decomposition, we start from arbitrary bases for A and B : then | Ψ AB i = a jk | j A i| k B i , (8) where j and k are summed over, and a is not a square matrix unless A and B happen to have the same dimensionality. The “singular value decomposition” of a jk implies | Ψ AB i = u ji d ii v ik | j A i| k B i , (9) where d is a (again, not generally square) diagonal matrix with real elements, and u and v are unitary matrices. We identify λ i = d ii , u
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

Page1 / 5

p212ln29 - Physics 212 Statistical mechanics II Fall 2006...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online