Physicists have long studied Landau states and energy levels of electrons in a stable magnetic field. However, when the magnetic field dissipates, the question of how Landau states evolve remains unanswered. This study investigates the magnetic dissipative effect from a quantum mechanics perspective. We establish a quantum master equation that describes a dissipative magnetic field. Subsequently, we solve the master equation using an entangled state representation. Consequently, we found that the Landau state evolves into a binomial quantum state (mixed state) when the magnetic field dissipated. This is in sharp contrast to the case where the Landau state becomes a negative binomial state when an additional harmonic oscillator potential is applied. We hope that this finding will be helpful for the evolution of quantized plasma physics.

The Lorentz force influences the motion of the charged particles in a magnetic field. Landau was the first to consider the quantum effect of electrons in a stable, uniform magnetic field. Landau discovered a discrete energy state called the Landau state.1 An interesting question that has long been overlooked is how the Landau state dissipates when the magnetic field is damped, which differs from the situation in which an additional harmonic potential is applied. In this study, we addressed these two questions.

Maxwell’s equations indicate that a time-varying magnetic field is accompanied by a spatially-varying electric field. The E field associated with the time-dependent B field satisfies
$∇×E=−∂B∂t.$
(1)
How does a dissipative magnetic field affect the static Landau state? In this study, we address this question using quantum mechanical concepts. In Ref. 2, the electron’s position is described via the coordinate eigenstates $λ$,
$λ=exp−12λ2−iλπ++iπ+K++λ*K+00,λ=λ1+iλ2,$
(2)
$π±=12MΩπx±iπy,π−,π+=1.$
(3)
Here, Ω = eB/M refers to the synchronous rotational frequency of electrons moving in a magnetic field, and πx and πy are the canonical momenta (instead of the mechanical momentum), given by
$πx=px+eAx,πy=py+eAy.$
(4)
The vector potential A is related to the magnetic field B (in the z direction) by
$A=−12By,12Bx,0,B=B.$
(5)
Then we explain K+ in Eq. (2) as
$K±=MΩ2x0∓iy0,$
(6)
where x0 and y0 are another pair of dynamic variables, called the guiding center by Johnson and Lippmann in Ref. 3,
$x0=x−πyMΩ,y0=y+πxMΩ.$
(7)
x0 and y0 cannot be accurately measured simultaneously because their commutation relationship is
$x0,y0=iMΩ.$
(8)
Other commutative relationships between those operators are
$π−,π+=1,K−,K+=1,πx,πy=−iMΩ,K±,π±=0,x0,π±=1,y0,π±=0,x,y=0.$
(9)
Then the Hamilton H describing this system is expressed as (setting = c = 1)
$H=12Mp+eA2=12Mπx2+πy2=π+π−+12Ω.$
(10)
Since the vacuum state $00$ is annihilated by π and K, i.e.,
$π−00=0,K−00=0,$
(11)
from Eq. (2), we have
$K++iπ−|λ〉=λ|λ〉,K−−iπ+|λ〉=λ*|λ〉,$
(12)
and we see that |λ⟩ is an entangled state representing the electronic coordinate’s eigenstate,
$x|λ〉=2MΩλ1|λ〉,y|λ〉=−2MΩλ2|λ〉.$
(13)
One can further prove that |λ⟩ is orthonormal and complete,2 so |λ⟩ makes up the electron’s coordinate representation in the presence of magnetic field. It should be noted that because of the presence of a magnetic field, the electrons are not in free motion.
The form of |λ⟩ is characteristic of the properties of entangled states, revealing the entanglement between the electromagnetic field and electrons. The eigenstates of π+π and K+K are as follows:
$n,m=π+nK+mn!m!00=π+nn!0,mn,m=0,1,2,….$
(14)
From Eq. (10), we see
$Hn,n−ml=n+12Ωn,n−ml.$
(15)
Let Lz = xpyypx. From Eqs. (4), (7), and (6), we obtain
$Lz=x0+πyMΩπy−eAy−y0−πxMΩπx−eAx=π+π−−K+K−,Lzn,n−ml=mln,n−ml,$
(16)
where ml is the angular momentum quantum number; therefore, $n,n−ml$ is a Landau state with definite energy and angular momentum. The following formula can be derived easily:
$π+n,m=n+1n+1,m,π−n,m=nn−1,m,$
(17)
so π+ and π can be seen as creation and annihilation operators, respectively.
When the magnetic amplitude dissipates, the Landau state (pure state) evolves into a mixed state, which is denoted as the density operator ρ. In this section, with the accompanying classical magnetic field amplitude varying, we shall introduce the quantum master equation of ρ and then solve it. For this purpose, we introduced eigenstates for π,
$π−ζ=ζζ,$
(18)
where $ζ$ belongs to the coherent states
$ζ=expζπ+−ζ*π−0,π−0=0$
(19)
and satisfies
$∫d2ζπζζ=1.$
(20)
Thus, any density operator ρ can be expanded by utilizing the completeness of the coherent states, that is, $Pζ,t$, where
$ρt=∫d2ζπPζ,tζζ,$
(21)
where the classical function $Pζ,t$ is a P representation of $ρt$. First, the amplitude decay of the coherent state is discussed,
$ζζ→ζe−κtζe−κt,$
(22)
where κ is the decay rate. Using the normal product form of $00=:exp(−π+π−):$, we rewrite
$ζe−κtζe−κt=:exp−ζ2e−2κt+ζe−κtπ++ζ*e−κtπ−−π+π−.$
(23)
It then follows
$ddtζe−κtζe−κt=ddt:exp−ζ2e−2κt+ζe−κtπ++ζ*e−κtπ−−π+π−:=2κζ2e−2κtζe−κtζe−κt−κπ+ζe−κtζe−κtζe−κt−κζe−κtζe−κtζ*e−κtπ−=2κπ−ζe−κtζe−κtπ+−κπ+π−ζe−κtζe−κt−κζe−κtζe−κtπ+π−.$
(24)
Let $ζe−κtζe−κt=ρt$. Then Eq. (24) is equivalent to
$dρdt=κ2π−ρπ+−π+π−ρ−ρπ+π−.$
(25)
Because we can expand any density operator in the coherent state representation, Eq. (25) is the master equation for the density operator, which represents the dissipated magnetic field.
In order to solve Eq. (25), we introduce another entangled state,
$η=exp−12η2+ηπ+−η*π̃++π+π̃+00̃.$
(26)
Eq. (25) differs from Eq. (2), where $π̃+$ is a fictitious mode accompanying the real mode π+, $π̃−,π̃+=1$, $π̃−0̃=0$, and $η$ obeys the eigenvector equations
$π−−π̃+|η〉=η|η〉,π+−π̃−|η〉=η*|η〉,ηπ+−π̃−=η*η,ηπ−−π̃+=ηη.$
(27)
Using the normal ordering form of the vacuum projector
$00̃00̃=:e−π+π−−π̃+π̃−:$
(28)
and the technique of Integration Within an Ordered Product (IWOP)4–8 of the operator, we can show the orthogonality as
$η′η=πδη′−ηδη′*−η*,$
(29)
as well as the completeness relation
$∫d2ηπηη=∫d2ηπ:×exp−|η|2+ηπ+−η*π̃++η*π−−ηπ̃−+π+π̃++π−π̃−−π+π−−π̃+π̃−:=1.$
(30)
Here we employed integration formula,9
$∫d2⁡απexp−hα+sα+gα=1hexpsgh,Reh<0.$
(31)
Let
$η=0=eπ+π̃+00̃≡I.$
(32)
We have
$π−I=π̃+I,π+I=π̃−I,(π+π−)nI=(π̃+π̃−)nI.$
(33)
Acting on both sides of Eq. (25) on the state $I$ and letting $ρ=ρI$, we get
$ddtρ=κ2π−ρπ+−π+π−ρ−ρπ+π−I.$
(34)
Using Eq. (33) and noting that the density operators defined in real space are commutative with those in tilde space, Eq. (34) becomes
$ddtρ=κ2π−π̃−−π+π−−π̃+π̃−ρ,$
(35)
so its formal solution is
$ρ=expκt2π−π̃−−π+π−−π̃+π̃−ρ0,$
(36)
where $ρ0≡ρ0I$, with ρ0 being the initial density operator.
It is noteworthy that the operators in Eq. (35) obey the following commutative relations:
$π−π̃−,π+π−=π−π̃−,π̃+π̃−=π̃−π−,π+π−+π̃+π̃−2,π−π̃−=−π̃−π−.$
(37)
Using the operator identity10
$eλA+σB=eλA⁡expσ1−e−λτB/τ$
(38)
(valid for [A, B] = τB), we have
$expκt2π−π̃−−π+π−−π̃+π̃−=exp−κtπ+π−+π̃+π̃−expTπ−π̃−,$
(39)
where
$T=1−e−2κt.$
(40)
Then substituting Eq. (39) into Eq. (36) and using Eq. (33) yield
$ρ=exp−κtπ+π−+π̃+π̃−∑j=0∞Tjj!π−jπ̃−jρ0=e−κtπ+π−∑j=0∞Tjj!π−jρ0π+je−κtπ̃+π̃−I=∑j=0∞Tjj!e−κtπ+π−π−jρ0π+je−κtπ+π−I,$
(41)
$ρt=∑j=0∞Tjj!e−κtπ+π−π−jρ0π+je−κtπ+π−=∑j=0∞Mjρ0Mj†,$
(42)
where
$Mj=Tjj!e−κtπ+π−π−j,$
(43)
satisfying $∑j=0∞MjMj†=1$, which is trace conservative $Trρt=1$ (see the  Appendix). Therefore, $ρt$ in Eq. (42) qualifies as a density operator.
We now consider how Landau states dissipate. In Eq. (42), let the initial state $ρ0=n,mn,m,$ and $n,m=π+nn!0,m$. Substituting ρ0 into Eq. (42) and using
$π−jn,m=n!n−j!n−j,m,$
(44)
we obtain
$ρt=∑j=0∞Tjj!e−κtπ+π−π−jn,mn,mπ+je−κtπ+π−=∑j=0nTjj!e−κtπ+π−n!n−j!n−j,mn−j,me−κtπ+π−=∑j=0nTjj!e−2κtn−jn!n−j!n−j,mn−j,m=∑j=0n1−e−2κtjCnje−2κtn−jn−j,mn−j,m=∑j′=0nCnn−j′e−2κtj′1−e−2κtn−j′j′,mj′,m.$
(45)
Noticing $Cnn−j′=Cnj′$, we have
$ρt=∑j′=0nCnj′e−2κtj′1−e−2κtn−j′j′,mj′,m.$
(46)
Therefore, we conclude that $ρt$ is a binomial state (a mixed state) where e−2κt is the binomial parameter and the initial pure Landau state evolves into a binomial state. The binomial state interpolates between the coherent state and the Landau state.11

In sharp contrast to the case where the Landau state decays in a magnetic field and degenerates into a binomial state (mixed state), the Landau state becomes a negative binomial state when an additional harmonic oscillator potential is applied.

When an electron in a uniform magnetic field is subjected to a harmonic oscillator potential, the Hamiltonian is
$H=12Mπx2+πy2+12Mω2x2+y2.$
(47)
Using Eqs. (3), (6), and (7), we can rewrite H as
$H=Gπ+π−+12+FK+K−−iFK+π+−K−π−+12F,$
(48)
where G = Ω + ω2/Ω and F = ω2/Ω. Employing the squeezing operator expressed in the entangled state representation,
$S=∫d2λπγλγλ,γ=ΩΩ′1/2,Ω′=Ω2+4ω2,$
(49)
since $Sλ=1γλγ$, we can diagonalize H,
$SHS−1=12Ω+Ω′π+π−+Ω−Ω′K+K−+Ω′.$
(50)
Hence, the Landau state becomes $S−1n,0$; thus we can derive
$S−1n,0=∑n=0∞Cn+llsec⁡hfn+1−i⁡tanh⁡fln+l,l.$
(51)
Here, $f=14lnΩ2Ω2+4ω2$; this is a negative binomial state.

In summary, quantum mechanical techniques were used for the first time to demonstrate the impact of a dissipative magnetic field on the Landau state. By establishing and solving a quantum master equation to dissipate the magnetic field, we found that the pure Landau state evolves into a binomial quantum state (mixed state) when the magnetic field dissipates. This is in sharp contrast to the case where the Landau state becomes a negative binomial state when an additional harmonic oscillator potential is applied. We hope that this finding will help further advance the relevant work in quantized plasma physics.

This work was sponsored by the Anhui Provincial Natural Science Foundation (Grant No. 2008085MA22) and the Natural Science Foundation of the Anhui Higher Education Institutions of China (Grant No. KJ2020A0673).

The authors have no conflicts to disclose.

Cheng Da: Conceptualization (lead); Formal analysis (lead); Funding acquisition (lead); Investigation (lead); Methodology (lead); Project administration (lead); Resources (lead); Supervision (equal); Validation (equal); Writing – original draft (lead); Writing – review & editing (equal). Hong-Yi Fan: Conceptualization (supporting); Investigation (supporting); Methodology (supporting); Supervision (equal); Validation (equal); Writing – original draft (supporting); Writing – review & editing (equal).

The data that support the findings of this study are available within the article.

Here, we first prove $∑j=0∞MjMj†=1$ for Mj in Eq. (43).

Using formula12
$:eef−1π+π−:=efπ+π−,$
(A1)
we can prove
$∑j=0∞Mj†Mj=∑j=0∞Tjj!π+je−2κtπ+π−π−j=∑j=0∞Tjj!e2jκt:π+jπ−j:e−2κtπ+π−=:eTe2κtπ+π−:e−2κtπ+π−=1.$
(A2)
Then we have
$Trρt=Tr∑n=0∞Mnρ0Mn†=Trρ0=1.$
(A3)
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