Quantum Optics Lecture 5 Summer Term 2019 Andrey Surzhykov Robert M¨uller Coherent states As we have seen and you will still proove number states do not have a well defined electric field. Thus, number states are not suitable to describe the quantized version of a classical field E(r; t). Can we find another quantum state that corresponds to E(r; t) in the classical limit? The intuitive way to achieve that would be a state with a huge number of photons N 1. However, as we have shown, the fact that hnjE^(r; t)jni = 0 does not depend on the photon number. Instead, since E^ depends on a sum ofa ^ anda ^y, we could construct a state that consists of Fock states with numbers jni and jn + 1i. Definition: Coherent states jαi are eigenfunctions of the annihilation operatora ^, so that: a^ jαi = α jαi (1) Because Fock states are a complete ONS, we can expand a coherent state in terms of number states, so that: 1 X jαi = cn jni : (2) n=0 If we act with the annihilation operator on this expansion, we get: 1 1 X X p a^ jαi = cna^ jni = ncn jn − 1i : (3) n=0 n=1 Because jαi is an eigenfunction ofa ^ we find the condition: 1 1 1 X p X X ncn jn − 1i = α cn jni = α cn−1 jn − 1i : (4) n=1 n=0 n=1 By comparison of the coefficients of the same Fock state, we find the recurrence relation: p ncn = αcn−1; (5) which we can rewrite: αn cn = c0 p (6) n! The only free parameter is the first coefficient c0, which can be fixed by normalization: 1 1 ∗ n n0 1 2 n 2 X X (α ) α 0 2 X (jαj ) 2 jαj2 ! hαjαi = jc0j p hnjn i = jc0j = jc0j e = 1 (7) 0 n! n=0 n0=0 n!n ! n=0 1 2 − jαj and therefore c0 = e 2 . Thus, a coherent state is constructed by: 1 1 2 n 2 n y n − jαj X α − jαj X α (^a ) jαi = e 2 p jni = e 2 p j0i (8) n=0 n! n=0 n! No let us show, that any state can be expanded in coherent states. Therefore we write: Z X Z dα jαi hαj = dα jni hnjαi hαjmi hmj n;m Z X jni hmj 2 = p dα e−|αj αnα∗m n!m! n;m | {z } =πn!δnm X = π jni hnj n Therefore we have found that the unity operator in terms of coherent states is: 1 Z dα jαi hαj = I^ (9) π The electric field of a coherent state The aim of the whole operation above was to construct a state with the proper classical limit. So let us calculate the expectation value of the electric field operator: r ! hαjE^(r; t)jαi = i ~ hαja^ ei(r·k−!t) − a^y e−i(r·k−!t)jαi 2"0V r ! = i ~ α ei(r·k−!t) − α∗ e−i(r·k−!t) 2" V 0 (10) r ! = i ~ jαj ei(r·k−!t+φ) − e−i(r·k−!t+φ) 2"0V r ! = −2 ~ jαj sin(r · k − !t + φ); 2"0V which corresponds to the classical electrical field. In the exercises you will show, that the variance of the electric field is ! hαj∆E^2(r; t)jαi = ~ : (11) 2"0V This corresponds to the variance of the vacuum and thus shows the definition of coherent states as states with minimal uncertainty. 2 Figure 1: A Fock or number state in phase space corresponds to a circular probability q 1 distribution around zero. The variance has a radius of n + 2 . Photon statistics of coherent states Since we can expand coherent states into number states, it is a straightforward question to ask, what the expectation value of the photon number is in such a state. As you will show it corresponds to: hαjn^jαi = jαj2 ≡ n (12) We can use this relation to find an expression for the probability distribution of the photon number. The probability to find n photons in a coherent state jαi can be calculated as follows: 2 2 m 2 − jαj X α 2 Pn = j hnjαi j = e p hnjmi m=0 m! 2n 2 jαj = e−|αj n! nn = e−n n! This corresponds to a Poisson-distribution! Fock and coherent states in phase space As we know from classical mechanics, each configuration at point r with momentum p can be represented by a point (r; p) in phase space. We can perform a similar consider- 3 ation in quantum optics. Let 1 r ! η^ = a^ +a ^y 1 2 1 p ! η^ = a^ − a^y 2 2i From these conjugated operatorsη ^1 andη ^2 we can build a 2D-phase space associated with a single-mode field. The expectation values of both operators with respect to a Fock state are hnjη^1=2jni = 0: The variance, however, is non-vanishing: 1 hnj∆^η2 jni = (2n + 1): 1=2 4 Sinceη ^1 andη ^2 do not commute we find the incertainty: 1 ∆^η2 ∆^η2 ≥ 1 2 16 Therefore we find for a number state in phase space the following relation: 1 ∆^η2 + ∆^η2 = n + (13) 1 2 2 q 1 which describes a circle with radius n + 2 , as depicted in Fig. 1. The description of a coherent state in phase space requires a bit more work. For the expectation values we find the interesting relation: 1 ∗ hαjη^1jαi = (α + α ) = Re(α) 2 (14) 1 hαjη^ jαi = (α − α∗) = Im(α) 2 2i This means that we can map the complex α-space to the phase space of a coherent state. Let us, therefore, use the polar representation of α: α = jαj eiθ The uncertaintyp of the radial distance jαj corresponds to the standard deviation of the photon number n. Becauseη ^1 andη ^2, again, do not commute, there is also a phase uncertainty. The variance of both operators is given by 1 ∆^η2 = ; (15) 1=2 4 which is the same as the vacuum uncertainty, which again shows that (i) coherent states are states of minimal uncertainty and (ii) as seen in Fig. 2 coherent states can be viewed 4 Figure 2: A coherent state jαi in phase space has a circular probability distribution. The center of this distribution is at (Re(α); Im(α). The 1 − σ-radius is the same as for the vacuum p1 . 2 as vacuum states displaced in phase space. Indeed we can construct an operator D^(α) that, acted on the vacuum state, performs exactly that shift, i.e.: D^(α) j0i = jαi (16) From Eq. (8) one might think that D^(α) = exp(−|αj2=2) exp(αa^y). This, however, is not true, because this exponential operator would not be unitary. Instead we define: D^(α) = eαa^y−α∗a^ (17) With the Baker-Campbell-Hausdorff formula A^+B^ 1 [A;^ B^] B^ A^ e = e 2 e e (18) and A^ = α∗a^ and B^ = αa^y we see that this corresponds to 2 − jαj αa^y α∗a^ D^(α) = e 2 e e ; (19) where we note that the exponent of the annihilation operator acted on the vacuum state has no effect and, thus, we retain Eq. (8) and, thus, have obtaines the unitary shifting operator we were looking for. Squeezed states In the previous section both phase-space variables of a coherent state had the same variance: 1 ∆^η2 = : (20) 1=2 4 5 Figure 3: A squeezed state jα; ξi in phase space has an asymmetric probability distribu- tion. The center of this distribution is at (Re(α); Im(α). The 1 − σ-distance from the center is modified by a factor of e−|ξj or ejξj, respectively. For various applications, however, it may be useful to construct states that have an 1 uncertainty smaller than 4 with respect to one obervable, while it is not important if another observable is defined less sharply. Therefore we aim to construct states, where: 1 ∆^η2 = γ 1 4 1 (21) ∆^η2 = ; 2 4γ 2 2 1 where γ ≤ 1. So that the product of both still satisfies h∆^η1i h∆^η2i ≥ 16 . With a complex number ξ we can define the squeezing operator: h 2i 1 ξ∗a^2−ξ(a^y) S^(ξ) = e 2 ; (22) which is a nonlinear operator ina ^ anda ^y. For consistency this operator has to be unitary: S^y(ξ) = S^(−ξ) S^y(ξ)S^(ξ) = S^(ξ)S^y(ξ) = I:^ Since we have seen that a coherent state is just a displaced vacuum state, we can for sim- plicity study the action of this operator on the quantum vacuum to generate a squeezed vacuum state: jξi = S^(ξ) j0i (23) 6 Now let us proof that those states are, indeed, squeezed states. Therefore let us calculate the expectation values: 1 ^y y ^ hξjη^1jξi = h0jS (ξ)(^a +a ^ )S(ξ)j0i 2 (24) 1 hξjη^ jξi = h0jS^y(ξ)(^a − a^y)S^(ξ)j0i 2 2i This, however, leaves us with the problem to calculate the operators S^y(ξ)^aS^(ξ) and S^y(ξ)^ayS^(ξ). In order to do this we can use the decomposition of exponential operators: 1 h i 1 h h ii eA^B^ e−A^ = B^ + [A;^ B^] + A;^ [A;^ B^] + A^ A;^ [A;^ B^] + ::: (25) 2! 3! h i ^ ^ y ^ 1 ∗ 2 y2 With B =a ^ or B =a ^ , respectively and A = 2 ξ a^ − ξ a^ we find: S^y(ξ)^aS^(ξ) =a ^ + ξa^y S^y(ξ)^ayS^(ξ) =a ^y + ξ∗a:^ Therefore: hξjη^1=2jξi = 0 (26) and, following analogue considerations: 2 1 −2r hξj∆^η1jξi = e 4 (27) 1 hξj∆^η2jξi = e2r; 2 4 where we have assumed that arg(ξ) = 0 and r is the so-called squeezing parameter.