{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

582-chapter8 - 8 Coherent State Path Integral Quantization...

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

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 8 Coherent State Path Integral Quantization of Quantum Field Theory 8.1 Coherent states and path integral quantization. 8.1.1 Coherent States Let us consider a Hilbert space spanned by a complete set of harmonic oscillator states {| n )} , with n = 0 ,..., ∞ . Let ˆ a † and ˆ a be a pair of creation and annihi- lation operators acting on that Hilbert space, and satisfying the commutation relations bracketleftbig ˆ a, ˆ a † bracketrightbig = 1 , bracketleftbig ˆ a † , ˆ a † bracketrightbig = 0 , [ˆ a, ˆ a ] = 0 (1) These operators generate the harmonic oscillators states {| n )} in the usual way, | n ) = 1 √ n ! ( ˆ a † ) n | ) (2) ˆ a | ) = 0 (3) where | ) is the vacuum state of the oscillator. Let us denote by | z ) the coherent state | z ) = e z ˆ a † | ) (4) ( z | = ( | e ¯ z ˆ a (5) where z is an arbitrary complex number and ¯ z is the complex conjugate. The coherent state | z ) has the defining property of being a wave packet with opti- mal spread, i.e., the Heisenberg uncertainty inequality is an equality for these coherent states. How does ˆ a act on the coherent state | z ) ? ˆ a | z ) = ∞ summationdisplay n =0 z n n ! ˆ a ( ˆ a † ) n | ) (6) Since bracketleftBig ˆ a, ( ˆ a † ) n bracketrightBig = n ( ˆ a † ) n − 1 (7) we get ˆ a | z ) = ∞ summationdisplay n =0 z n n ! parenleftBigbracketleftBig ˆ a, ( ˆ a † ) n bracketrightBig + ( ˆ a † ) n ˆ a parenrightBig | ) (8) Thus, we find ˆ a | z ) = ∞ summationdisplay n =0 z n n ! n ( ˆ a † ) n − 1 | ) ≡ z | z ) (9) Therefore | z ) is a right eigenvector of ˆ a and z is the (right) eigenvalue. 1 Likewise we get ˆ a † | z ) = ˆ a † ∞ summationdisplay n =0 z n n ! ( ˆ a † ) n | ) = ∞ summationdisplay n =0 z n n ! ( ˆ a † ) n +1 | ) = ∞ summationdisplay n =0 ( n + 1) z n ( n + 1)! ( ˆ a † ) n +1 | ) = ∞ summationdisplay n =1 n z n − 1 n ! ( ˆ a † ) n | ) (10) Thus, ˆ a † | z ) = ∂ ∂z | z ) (11) Another quantity of interest is the overlap of two coherent states, ( z | z ′ ) , ( z | z ′ ) = ( | e ¯ z ˆ a e z ′ ˆ a † | ) (12) We will calculate this matrix element using the Baker-Hausdorff formulas e ˆ A e ˆ B = e ˆ A + ˆ B + 1 2 bracketleftBig ˆ A, ˆ B bracketrightBig = e bracketleftBig ˆ A, ˆ B bracketrightBig e ˆ B e ˆ A (13) which holds provided the commutator bracketleftBig ˆ A, ˆ B bracketrightBig is a c-number, i.e., it is propor- tional to the identity operator. Since bracketleftbig ˆ a, ˆ a † bracketrightbig = 1, we find ( z | z ′ ) = e ¯ zz ′ ( | e z ′ ˆ a † e ¯ z ˆ a | ) (14) But e ¯ z ˆ a | ) = | ) (15) and ( | e z ′ ˆ a † = ( | (16) Hence we get ( z | z ′ ) = e ¯ zz ′ (17) An arbitrary state | ψ ) of this Hilbert space can be expanded in the harmonic oscillator basis states {| n )} , | ψ ) = ∞ summationdisplay n =0 ψ n √ n !...
View Full Document

{[ snackBarMessage ]}

Page1 / 36

582-chapter8 - 8 Coherent State Path Integral Quantization...

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

View Full Document
Ask a homework question - tutors are online