9512.net
甜梦文库
当前位置:首页 >> >>

A comment on W. Greiner's Quantum Mechanics -

A comment on W. Greiner’s “Quantum Mechanics – An Introduction”
J. -M. Chung

arXiv:physics/0306119v1 [physics.ed-ph] 15 Jun 2003

Research Institute for Basic Sciences and Department of Physics, Kyung Hee University, Seoul 130-701, Korea

Abstract
It is pointed out that in Greiner’s book, “Quantum Mechanics –An Introduction,” there is a confusion between the wavelength spectral function and the (angular) frequency spectral function in the derivation of the Wien’s displacement law.

In his book, Quantum Mechanics – An Introduction [1], Greiner derives Wien’s displacement law from Planck’s spectral energy density distribution. In his derivation, there is a mistake due to a confusion between the wavelength spectral function and the (angular) frequency spectral function. For convenience of discussion, here we quote the problem posed and a part of the solution to it. Problem. Derive Wien’s displacement law, i.e., λmax T = const.
1 from Planck’s spectral energy density V dE/dω. Here λmax is the wavelength 1 where V dE/dω achieves its maximum. Interpret the result.

Solution. We are looking for the maximum of the Planck’s spetral distribution:
?1 d hω 3 ? d 1 dE hω ? = ?1 exp dω V dω dω π 2 c3 kB T 2 ?1 3? ω h hω ? = 2 3 exp ?1 π c kB T 3 exp(? ω/kB T ) h ? hω h ? =0 ? 2 3 π c kB T [exp(? ω/kB T ) ? 1]2 h hω ? hω ? hω ? ?3? exp ?1 exp kB T kB T kB T

?1

= 0.

(1)

With the shorthand notation x =

h ?ω , kB T

we get the transcendental equation

1

ex = 1 ?

x 3

?1

,

(2)

which must be solved graphically or numerically. Besides the trivial solution x = 0 (minimum), a positive solution exists. Therefore xmax = hωmax ? , kB T (3)

and because ωmax = 2πνmax = 2πc/λmax we have λmax T = const. = 0.29 cm K . This means · · · · · · (4)

The Planck’s spectral distribution used in the above solution hω 3 ? hω ? 1 dE(ω, T ) = 2 3 exp V dω π c kB T
?1

?1

≡ u(ω, T ) ?

(5)

is the angular frequency spectral function. The wavelength and frequency spectral functions take the following forms: u(λ, T ) = 8πhc ch exp 5 λ kB λT 8πhν 3 hν u(ν, T ) = ? exp 3 c kB T
?1

?1
?1

, .

(6)

?1

Physically, these three spectral functions are related by the following equation: u(λ, T )dλ = ??(ν, T )dν = ??(ω, T )dω . u u Let us de?ne λM , νM , and ωM as the wavelength, frequency, and angular frequency at which u(λ, T ), u(ν, T ), and u(ω, T ) have their maximum value. (At this stage, it is very ? ? clear that the ‘ωmax ’ in Eq. (3) is the same as ωM .) For λM , one must solve the following transendental equation ex = 1 ? x 5
?1

,

(7)

ch with x = kB λT . The equation to solve for νmax is the same as the one in Eq. (2), this time, with x = khνT . The solutions to Eqs. (2) and (7) (x1 and x2 , respectively) are given as B follows:

x1 = 2.281 · · · , x2 = 4.965 · · · . 2

(8) (9)

Eqs. (8) and (9) yield 2πc T cT ch = = = 0.510 cm K , ωM νM kB x1 ch = 0.289 cm K . λM T = kB x2 Thus, we have ωM = 2πνM = 2.281 × 2πc/λM . 4.965

If the ‘λmax ’ in the underlined sentence (in the quoted problem) is de?ned as in the boxed equation above Eq. (4), in conformity with the following general relation: ω = 2πν = 2πc/λ , then, we have λmax T = 2.177 λM T = 0.629 cm K , (10)

which di?ers from Eq. (4). In conclusion, if the author of [1] insists that the ‘λmax ’ in Eq. (4) be the wavelength at which the wavelength spectral function u(λ, T ) achieves its maximum, in order to keep the number 0.29 on the right-hand side of Eq. (4) as usual [2], then the wavelength spectral function u(λ, T ) of Eq. (6), instead of the (angular) frequency spectral function, should be used.

3

REFERENCES
[1] W. Greiner, Quantum Mechanics — An Introduction, 4th ed. (Springer, Berlin, 2001), p. 24. [2] For example, R. L. Libo?, Introductory Quantum Mechanics, 3rd. ed. (Addison-Wesley, 1998), Problem 2.4 on p. 35.

4



学霸百科 | 新词新语

All rights reserved Powered by 甜梦文库 9512.net

copyright ©right 2010-2021。
甜梦文库内容来自网络,如有侵犯请联系客服。zhit325@126.com|网站地图