Abstract

The method of partial distributions is used to calculate the population distribution of three level maser systems in which spontaneous emission and cross relaxation as well as thermal relaxation and induced absorption processes are included. The results obtained are in agreement with the solution of the relevant rate equations. The method has the advantage of speed and suggestiveness. Two examples are presented: (1) A three level optical maser is analyzed in which one spontaneous emission transition probability is included. The solution in this example does not make the linear approximation to the Boltzmann factor. (2) A three level microwave maser is analyzed in which one cross relaxation mechanism is included. The solution to this problem by the method of partial distributions does require use of the linear approximation to the Boltzmann factor.

© 1962 Optical Society of America

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References

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  1. J. D. Keating, L. E. Follis, W. A. Barker, Bull. Am. Phys. Soc. Ser. II, 6, 68 (1961).
  2. T. Maiman, Nature 187, 493 (1960); Phys. Rev. Letters 4, 564 (1960).
    [CrossRef]
  3. N. Bloembergen, S. Shapiro, P. S. Pershan, J. O. Altman, Phys. Rev. 114, 445 (1959).
    [CrossRef]

1961

J. D. Keating, L. E. Follis, W. A. Barker, Bull. Am. Phys. Soc. Ser. II, 6, 68 (1961).

1960

T. Maiman, Nature 187, 493 (1960); Phys. Rev. Letters 4, 564 (1960).
[CrossRef]

1959

N. Bloembergen, S. Shapiro, P. S. Pershan, J. O. Altman, Phys. Rev. 114, 445 (1959).
[CrossRef]

Altman, J. O.

N. Bloembergen, S. Shapiro, P. S. Pershan, J. O. Altman, Phys. Rev. 114, 445 (1959).
[CrossRef]

Barker, W. A.

J. D. Keating, L. E. Follis, W. A. Barker, Bull. Am. Phys. Soc. Ser. II, 6, 68 (1961).

Bloembergen, N.

N. Bloembergen, S. Shapiro, P. S. Pershan, J. O. Altman, Phys. Rev. 114, 445 (1959).
[CrossRef]

Follis, L. E.

J. D. Keating, L. E. Follis, W. A. Barker, Bull. Am. Phys. Soc. Ser. II, 6, 68 (1961).

Keating, J. D.

J. D. Keating, L. E. Follis, W. A. Barker, Bull. Am. Phys. Soc. Ser. II, 6, 68 (1961).

Maiman, T.

T. Maiman, Nature 187, 493 (1960); Phys. Rev. Letters 4, 564 (1960).
[CrossRef]

Pershan, P. S.

N. Bloembergen, S. Shapiro, P. S. Pershan, J. O. Altman, Phys. Rev. 114, 445 (1959).
[CrossRef]

Shapiro, S.

N. Bloembergen, S. Shapiro, P. S. Pershan, J. O. Altman, Phys. Rev. 114, 445 (1959).
[CrossRef]

Bull. Am. Phys. Soc. Ser. II

J. D. Keating, L. E. Follis, W. A. Barker, Bull. Am. Phys. Soc. Ser. II, 6, 68 (1961).

Nature

T. Maiman, Nature 187, 493 (1960); Phys. Rev. Letters 4, 564 (1960).
[CrossRef]

Phys. Rev.

N. Bloembergen, S. Shapiro, P. S. Pershan, J. O. Altman, Phys. Rev. 114, 445 (1959).
[CrossRef]

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Figures (7)

Fig. 1
Fig. 1

Spontaneous emission in a three level optical maser.

Fig. 2
Fig. 2

Partial distribution due to W13 and w32.

Fig. 3
Fig. 3

Partial distribution due to W13 and A21.

Fig. 4
Fig. 4

Partial distribution due to W32 and A21.

Fig. 5
Fig. 5

Cross relaxation in a three-level microwave maser.

Fig. 6
Fig. 6

Partial distribution due to W13 and w1232.

Fig. 7
Fig. 7

Partial distribution due to W13 and w32.

Equations (15)

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ρ 1 k = ρ 3 k for k = 1,2.
ρ 2 1 ρ 3 1 = e δ = w 32 w 23 .
ρ 2 2 = ρ 2 3 = 0.
a i k = N ρ i k i ρ i k .
f k = X k i ρ i k i , k X k ρ i k .
f 1 = W 13 ( w 32 + 2 w 23 ) W 13 ( w 32 + 2 w 23 ) + 2 W 13 A 21 + w 32 A 21 , f 2 = 2 W 13 A 21 W 13 ( w 32 + 2 w 23 ) + 2 W 13 A 12 + w 32 A 21 , f 3 = w 32 A 21 W 13 ( w 32 + 2 w 23 ) + 2 W 13 A 21 + w 32 A 21 .
n i = k a i k f k .
n 1 = N [ W 13 ( w 23 + A 21 ) + w 32 A 21 W 13 ( w 32 + 2 w 23 ) + 2 W 13 A 21 + w 32 A 21 ] , n 2 = N [ W 13 w 32 W 13 ( w 32 + 2 w 23 ) + 2 W 13 A 21 + w 32 A 21 ] , n 3 = N [ W 13 ( w 23 + A 21 ) W 13 ( w 32 + 2 w 23 ) + 2 W 13 A 21 + w 32 A 21 ] .
n 1 + n 2 + n 3 = N
d n 1 d t = n 1 W 13 + n 3 W 31 + n 2 A 21 = 0 , d n 2 d t = n 2 A 21 n 2 w 23 + n 3 w 32 = 0 , d n 3 d t = n 3 W 31 + n 1 W 13 n 3 w 32 + n 2 w 23 = 0.
d n 1 d t = n 1 W 13 + n 3 W 31 W 1232 N ( n 1 n 3 n 2 2 ) = 0 , d n 2 d t = w 2123 N ( n 2 2 n 1 n 3 ) + n 3 w 32 n 2 w 23 = 0 , d n 3 d t = n 3 W 31 + n 1 W 13 + n 2 w 23 n 3 w 32 w 3212 N ( n 1 n 3 n 2 2 ) = 0 , n 1 + n 2 + n 3 = N ,
d n 2 d t = 2 w 1232 N ( n 2 n 1 ) ( n 2 + n 1 ) + n 1 w 32 n 2 w 23 4 3 w 1232 ( n 2 n 1 ) + n 1 w 32 n 2 w 23 = 0.
ρ 1 1 ρ 2 1 = ρ 2 1 ρ 3 1 .
f 1 = 4 w 1232 4 w 1232 + 3 w 32 f 2 = 3 w 32 4 w 1232 + 3 w 32 .
n 1 = n 3 = N 3 [ 1 h ν 32 k t w 32 4 w 1232 + 3 w 32 ] , n 2 = N 3 [ 1 + 2 h ν 32 k t w 32 4 w 1232 + 3 w 32 ] .

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