# QM Probability Current

View again

## Documents

8 views
PDF
All materials on our website are shared by users. If you have any questions about copyright issues, please report us to resolve them. We are always happy to assist you.
Document Description
QM Probability Current
Document Share
Document Tags

## Momentum

Document Transcript
Probability Current and Current Operators in Quantum Mechanics G. M. Wysinwysin@phys.ksu.edu, http://www.phys.ksu.edu/personal/wysinDepartment of Physics, Kansas State University, Manhattan, KS 66506-2601Written August, 2011, Vi¸cosa, BrazilSummary A quantum particle such as an electron produces electric current because of its motion.That current is associated with the ﬂow of its probability. The form of the wave functionthat describes the state of a particle determines these currents. At a more advancedlevel, one can ﬁnd quantum operators that can act between states, or work together witha density matrix, to deﬁne the currents even in a situation such as the mixed states of thermal equilibrium. The ideas and equations used to apply these ideas are summarizedhere. 1 Quantum Particle Motion One can consider quantum particles of charge  e , mass  m , momentum operator ˆ p , whose dynamicsis determined by a nonrelativistic Hamiltonian,ˆ H   = 12 m  ˆ p −  ec ˆ A (ˆ r ,t )  2 +  eφ (ˆ r ,t ) +  U  (ˆ r ) (1) c  is the speed of light, ˆ A  is the vector potential,  φ  is the scalar potential, and  U   is any other potential,such as that of a crystal lattice, that aﬀects the particles. For this problem, including applied EMﬁelds (which really are responsible for currents in many situations), it is convenient to see that theHamiltonian depends on a ”kinetic momentum” operator,  π , deﬁned by π  = ˆ p −  ec ˆ A .  (2)In the real space representation of QM, the momentum operator is ˆ p  =  − i ¯ h  ∇ , and the vectorpotential depends only on the space coordinates and time. The main topic of these notes is toﬁnd and describe the operations and operators that would be used to calculate the electric currentsassociated with the motion of these particles, assumed to be electrons 1 . That is a useful thing tohave, because it will allow for determining their induced electric and magnetic dipole moments,which exhibit themselves in the optical properties of materials.The main question to be addressed here is: What is the operator for the electric current density?Or, given a particular state of a quantum system, how does one ﬁnd the distribution of electriccurrent density? By the form of this question, mainly we are concerned with states described inreal space, so the interest is primarily in their real space representations, or their real space wavefunctions.Charge current is associated with the quantum motion of the charges. But motion in quantummechanics is probabilistic, hence, the motion one talks about is how the probability for ﬁnding theparticle moves aroud with time. So the main idea is that one needs to ﬁnd a ”probability current”that relates to how the probability for locating the electron might be changing with time, when awave function satisﬁes a Schrodinger equation based on the above Hamiltonian.In a semi-classical sense, we need to ﬁnd the eﬀective velocity operator ˆ v  or current densityoperator ˆ  j  for one quantum particle. The electric charge density  ρ e  for an individual electron needs 1 The charge will be written as  e , although for real electrons that is a negative number,  e  = − 4 . 8 × 10 − 10 esu, inCGS units, or  e  = − 1 . 602 × 10 − 19 coulombs, in SI units. These notes use CGS. 1  to satisfy a continuity equation that ensures conservation of charge (these are QM expectationvalues), ∂ ∂tρ e ( r ) +  ∇ ·  j ( r ) = 0 ,  or  ∇ ·  j  =  −  ˙ ρ e .  (3)Note that these quantities are associated with a single electron (but assume they have the correctunits). To get the physical charge and current densities, scale them by the number of electrons  N  , ρ Ne  =  Nρ e ,  J  =  N   j .  (4) 1.1 Current density in a wave function First, consider the usual elementary approach, based on properties of a given arbitrary wave function ψ ( r ,t ). This wave function could be an energy eigenstate of the Hamiltonian, or any mixture of thoseeigenstates, it really doesn’t matter. The wave function evolves according to a Schr¨odinger equation, i ¯ h  ˙ ψ  = ˆ Hψ , and its complex conjugate satisﬁes  − i ¯ h  ˙ ψ ∗ = ˆ H  ∗ ψ ∗ . The squared wave function givesthe probability density, so the charge density is deﬁned to be  ρ e  =  e | ψ | 2 . One can get the currentby looking at changes of   ρ e ( r ,t ) with time, which depend on changes in  | ψ | 2 : ∂ ∂t | ψ | 2 =  ∂ ∂tψ ∗ ψ  = ˙ ψ ∗ ψ  +  ψ ∗  ˙ ψ  = 1 − i ¯ h ( ˆ H  ∗ ψ ∗ ) ψ  + 1 i ¯ hψ ∗ ( ˆ Hψ ) (5)From the continuity equation, this can be connected to the current we are looking for, by −   ∇ ·  j  =  e ∂ ∂t | ψ | 2 .  (6)The Hamiltonian has the nonrelativistic form (1), where the potential  U  ( r ) is assumed to be real.The EM potentials  A  and  φ  also technically must be real, because they describe the real electricand magnetic ﬁelds. At some points in calculations they may be taken as complex, but always, itis assumed that such expressions have an implied extra complex conjugate term, to make the totalreal. Using this Hamiltonian, the potentials cancel out, and only the kinetic energy terms remain, e ∂ ∂t | ψ | 2 =  ei ¯ h 12 m  ψ ∗ ( π 2 ψ )  −  ( π 2 ψ ) ∗ ψ   (7)For simplicity of notation, and based on the momentum operator ˆ p  =  − i ¯ h  ∇ , I introduce the deﬁni-tion, π  =  − i ¯ h (   ∇ −  i α ) ,  where   α  =  e ¯ hc ˆ A .  (8)This gives e ∂ ∂t | ψ | 2 =  ei ¯ h ( − i ¯ h ) 2 2 m  ψ ∗  (   ∇ −  i α ) 2 ψ  −  (   ∇  +  i α ) 2 ψ ∗  ψ   (9)The scaled vector potential   α  also is a real function of space and time. When the kinetic momentumis inserted and squared out, the terms depending on   α 2 can be seen to cancel out. What remainsgives two diﬀerent contributions that produce currents. The ﬁrst terms depending on  ∇ 2 will givethe “momentum current”, so I place a subscript  p  to denote this part,   ∂ ∂te | ψ | 2   p =  ei ¯ h ( − i ¯ h ) 2 2 m  ψ ∗ ( ∇ 2 ψ )  −  ( ∇ 2 ψ ) ∗ ψ   =    ∇ ·  ie ¯ h 2 m  ψ ∗ (   ∇ ψ )  −  (   ∇ ψ ∗ ) ψ   (10)Via the continuity equation as in (6), the momentum current contribution (expectation value) is  j  p ( r ) =  − ie ¯ h 2 m  ψ ∗ (   ∇ ψ )  −  (   ∇ ψ ∗ ) ψ   =  e 2 m  [ ψ ∗ (ˆ p ψ ) + (ˆ p ψ ) ∗ ψ ] =  em Re { ψ ∗ (ˆ p ψ ) }  (11)This is the action on a wave function  ψ ( r ) =   r | ψ  . One sees that this depends on an operator e ˆ p /m . However, to apply this formula, the indicated product must be formed, and then the real2  part must be taken. There should be a slightly diﬀerent (and more complicated) method to inventan individual operator whose eﬀect between  ψ ∗ and  ψ  does not require the real part operation. Thatis considered later in these notes.The cross terms between    ∇  and   α  give a term which might be called the “gauge curent” part,   ∂ ∂te | ψ | 2  A =  i ¯ he 2 m  − iψ ∗  (   ∇ ·   α  +   α  ·    ∇ ) ψ  −  i  (   ∇ ·   α  +   α  ·    ∇ ) ψ ∗  ψ  =  e ¯ h 2 m  ψ ∗    ∇ ·  (  αψ ) +   α  ·  (   ∇ ψ )  +    ∇ ·  (  αψ ∗ ) +   α  ·  (   ∇ ψ ∗ )  ψ   (12)Note that the second term is the complex conjugate of the ﬁrst, so the result is absolutely real. If we assume the Coulomb gauge, then    ∇·    A  = 0 and    ∇·  α  = 0. The relation    ∇· (  αψ ) =   α ·    ∇ ψ  results,and this simpliﬁes to   ∂ ∂te | ψ | 2  A =  e ¯ h 2 m  2 ψ ∗  α  ·    ∇ ψ  + 2 ψ α  ·    ∇ ψ ∗   =    ∇ ·  ψ ∗  e 2 mc  A ψ   (13)Since    A  is real, this object is automatically real and so there is no need for a real part operation infront of it. Thus the gauge current density (expectation value) is  j A  =  −  e 2 mcψ ∗   A ψ  (14)and its operator is just  − e 2   A /mc . Therefore the expectation value of the total current density atpoint  r  in a wavefunction  ψ ( r ) is  j ( r ) =  j  p  +  j A  = Re   emψ ∗ ˆ p ψ  −  e 2 mcψ 2  ˆ A ψ  = Re   em ψ ∗ ( r )  ˆ p −  ec ˆ A  ψ ( r )  .  (15)One can realize we can write this in terms of a velocity operator; this could make more physicalsense:  j ( r ) = Re { eψ ∗ (ˆ v ψ ) } ,  ˆ v  =  πm  = 1 m  ˆ p −  ec ˆ A   (16)The general velocity operator depends on the total vector potential, that could be due to both ACoptical ﬁelds and a DC magnetic ﬁeld. We could have written this based on intuition but it is good tosee the mathematical description works within the Schr¨odinger equation. Unfortunately, the resultrequires the application of   real   at the end of the calculation. This is OK but somewhat annoying,because it does not ﬁt into the usual formalism for quantum expectation values. Note, however, thisobject is a density and not a matrix element. For a Hermitian operator like ˆ p , a matrix element,involving integrating over all space, will be real. But is it curious that the associated density couldbe complex. 1.1.1 Simple examples Suppose there is no vector potential, and the wave function is a  plane wave , ψ  =  C  0 e i ( kx − ωt ) (17)traveling in one dimension with wave vector  k  and frequency  ω . The value of   C  0  is determined bythe choice of volume of normalization. This would actually be stationary state of Hamiltonian (1) inthe absence of applied ﬁelds and potentials (free qm particle). It describes a particle with preciselydeﬁned momentum  p  = ¯ hk  and kinetic energy  E   = (¯ hk ) 2 / 2 m . So it should have a nonzero current.One has easily ˆ  p x ψ  = ¯ hkψ , and also then  ψ ∗ ˆ  p x ψ  = ¯ hk | ψ | 2 . This is already a real number. Thenthe current density according to (15) is  j x ( r ) =  emψ ∗ ˆ  p x ψ  =  em ¯ hk | C  0 | 2 .  (18)3  One could think of the velocity as  v x  =  p x /m  = ¯ hk/m , then this result is the same as  j x  =  e | C  0 | 2 v x .This is the current  density  , which is seen to be independent of position. Everywhere the plane waveis, there is a uniform ﬂow of charge (and probability), even though it is a stationary state! Note alsothat the charge density is  ρ e  =  ψ ∗ eψ  =  e | C  0 | 2 is also a uniform constant in space. In fact, if thewave function is normalized to unity over some volume  V   (the volume available to each electron)then the normalization constant is  C  0  = 1 / √  V   . So really, the result is the same as  j x ( r ) =  ρ e ¯ hkm  =  ρ e v x .  (19)This is a statement that the product of volume charge density and velocity gives the current density,which is totally the expected relationship.Next, suppose the wave function is a  linear combination of real and imaginary parts .Imagine we just have ψ  =  ψ 1  +  iψ 2  (20)where each of the individual functions are real. Then also including the vector potential, one gets ψ ∗ e ˆ v ψ  = ( ψ 1  −  iψ 2 )  em  − i ¯ h  ∇ −  ec ˆ A ( r )  ( ψ 1  +  iψ 2 )=  e ¯ hm ( ψ 1   ∇ ψ 2  −  ψ 2   ∇ ψ 1 )  −  e 2 mc ( ψ 21  +  ψ 22 ) A ( r )  −  i ¯ hem  ( ψ 1   ∇ ψ 1  +  ψ 2   ∇ ψ 2 ) (21)That’s a very interesting result. Of course, one needs to take the real part to actually get the physicalcurrent. So only the ﬁrst two terms contribute to electric current:  j ( r ) = Re { ψ ∗ e v ψ }  =  e ¯ hm ( ψ 1   ∇ ψ 2  −  ψ 2   ∇ ψ 1 )  −  e 2 A ( r ) mc  | ψ | 2 (22)The ﬁrst term, the momentum current, results from the interference between the real and imaginaryparts. It is zero in a wave function that is pure real or pure imaginary. This is usually the mostimportant part in most problems. So if a wave function does not have an ” i ” in it somewhere, themomentum current is zero. The second term is caused by the vector potential, and just depends onthe square of the wave function at the point of interest. In fact, it really depends on the electriccharge density,  ρ e  =  e | ψ | 2 . This is a curious term. The vector potential needed to describe someEM ﬁeld is not unique. However, remember that this expression is derived for the Coulomb gaugewhere    ∇· A  = 0. This is indeed a strong constraint, on the other hand, it still leaves a lot of freedomin the choice of   A . Thus one sees a gauge dependence, but what is not as apparent, is that themomentum current could also have a gauge dependence. The solutions for  ψ 1  and  ψ 2  will dependon the choice of   A . In any speciﬁed physical situation, in the end, the physically measured resultsshould come out independent of the choice of the EM gauge. 1.2 Doing quantum mechancis with the density matrix Here before really discussing currents, some general ideas about quantum mechanics for statisticalsystems are expanded upon. This usually will mean any real system, where there is more than oneparticle and there is lack of knowledge about the preparation of the system.When the state of a quantum system is speciﬁed by a single wave function, the system is saidto be in a  pure state  . There are certain physical variables that could be measured and one willdeﬁnitely know the probailities for each possible outcome. (Unfortunately, in QM, this is the bestthat one can hope to achieve!) This makes the most sense, for example, if that state is an eigenstateof the Hamiltonian. It also applies to any superposition of eigenstates of  ˆ H  , because we would beable to predict outcomes of experiments very precisely (in the sense of probabilities, however).But in the real world, one can never really prepare a system exactly how we want it. Almostalways, our system will be impossible to describe perfectly. We can say only that it is in a certaindesired state with some probability. It is as if we want to write the state as a superposition, but, we4
Similar documents

### Loi sur les lettres et billets de dépôt. Depository Bills and Notes Act. Current to January 31, 2017 À jour au 31 janvier 2017 CODIFICATION

View more...
Search Related

#### goquiolay

Thank you for visiting our website and your interest in our free products and services. We are nonprofit website to share and download documents. To the running of this website, we need your help to support us.

Thanks to everyone for your continued support.

No, Thanks
SAVE OUR EARTH

We need your sign to support Project to invent "SMART AND CONTROLLABLE REFLECTIVE BALLOONS" to cover the Sun and Save Our Earth.

More details...

Sign Now!

We are very appreciated for your Prompt Action!

x