Spectrum of Compatible Variables
Given three variablesA,B,C, demonstrate that if[A,B]=[A,C] =0, but[B,C] 0, the spectrum of Ais degenerate.
Suppose that all of the eigenvalues ofAare not degenerate, so that, for each eigenvalue aof A, there is only one ket|ψ a such that
If this were true, each ket|ψ a must also be eigenstate ofBandCthat are compatible withA As a consequence, we can also label the ket|ψ a with the eigenvalues ofB eC:
C|ψ a , b , c =c|ψ a , b , c where, obviously, onceais fixed,becmust be unique For each generic state|ψ, it results that
This contradicts our initial supposition that[B,C] =0. © Springer Nature Switzerland AG 2019 1
Constants of Motion
Show that, if FeGare two constants of motion for a quantum system, this is also true for[F,G].
IfFandGare two constants of motion, then, from the Heisemberg equation,
[G,H], whereHis the system Hamiltonian It turns out that d dt[F,G] =∂[F,G]
[FHG−HF G+F GH−FHG−GHF+HG F−G FH+GHF−
Hence,[F,G]is a constant of motion.
Number Operator
Let an operatorabe given that satisfies the following relationships: aa + +a + a=1, a 2 =(a + ) 2 =0.
(b) Prove that the only possible eigenvalues for operatorN =a + aare 0 and 1.
(a) Suppose thatais hermitian:a=a + We obtain aa + +a + a=2(a + ) 2 =0, which contradicts the initial statement.
It is widely recognized that if an operator fulfills an algebraic equation, its eigenvalues will also satisfy the same equation Specifically, by denoting |λ as the generic eigenket of the operator N corresponding to the eigenvalue λ, we can express this relationship clearly.
Momentum Expectation Value
Given a particle of massmin a potentialV( r ), system described by the Hamiltonian
Use this relationship to show that, in a stationary state, p =0.
Callingr i andp i (i=1,2,3)the position and momentum components, we have
2m(r i p i 2 −p i 2 r i ) = 1 2m(r i p i 2 −p i 2 r i −p i r i p i +p i r i p i ) = 1 2m([r i ,p i ]p i +p i [r i ,p i ]) = ip i m , as conjectured Calling|ψ E the eigenstate ofHcorresponding to an eigenvalueE, the expectation value of each momentum component is p i = ψ E |p i |ψ E = −im ψ E |[r i ,H]|ψ E = −im ψ E |r i E|ψ E − ψ E |Er i |ψ E
The validity of the result hinges on the well-defined nature of the quantity provided It is important to note that this result does not apply to improper eigenvectors, such as those associated with free particles, where |ψ E is treated as a simultaneous eigenstate of the Hamiltonian (H) and momentum (p), or more generally, as an eigenstate of the continuous spectrum.
Wave Function and the Hamiltonian
A particle is in a state described by the following wave function: ψ( r )= Asin pãr
(b) What can we say about the value of momentum and energy in this state?
(a) The wave function is representative of the dynamical state of a system To decide whether the particle is free, we need to know the Hamiltonian.
(b) We can write this wave function as ψ( r )= A
The article discusses the superposition of two momentum eigenstates, represented by eigenvalues +p and -p Since the coefficients in this linear superposition have equal magnitudes, the expectation value of momentum is zero However, without information about the Hamiltonian, no conclusions can be drawn regarding the energy.
What Does a Wave Function Tell Us?
A particle constrained to move in one dimension is described at a certain instant by the wave function ψ(x)=A coskx.
(a) it describes a state with defined momentum?
(b) it describes a free particle state?
(a) The wave function can be written as ψ(x)= A
It is the linear superposition of two momentum eigenstates with momentum p =k and p= −k As they have equal amplitudes, they are equiprobable.
So, the answer to the question is no The kinetic energy E= 2m p 2 is defined instead.
The wave function may serve as the eigenfunction of a Hamiltonian in a potential-free scenario, but it is crucial to note that it defines the state of a system rather than its dynamics The dynamics are determined by the Hamiltonian, which remains unspecified in this context.
Spectrum of a Hamiltonian
Consider a physical system described by the Hamiltonian
Find theαandβvalues for whichHis bounded from below and, if this is the case, find its eigenvalues and eigenvectors.
The Hamiltonian can be rewritten as:
• p is hermitian, being a linear combination of two hermitian operators, provided αis real.
(Note that these properties also apply in the case ofp =p+ f(q), with f(q)being a real function ofq.)
Impose thatHis bounded from below: ψ|H|ψ = 1
The first term being positive or zero, this condition is verified for every|ψ, provided β > mα 2
If this is true, we can replay the harmonic oscillator procedure for this Hamiltonian, obtaining the same eigenvalues and eigenstates In this case, the frequency is ω 2β m −α 2
Velocity Operator for a Charged Particle
Given a a charged particle in a magnetic field, find the commutation relations between the operators corresponding to the velocity components.
Remember that the Hamiltonian of a particle having chargeqin an electromagnetic field is
+qφ, whereAandφare the magnetic and electric potential giving rise to the electromag- netic field:B= ∇ ×A , E= − 1 c ∂ ∂ A t − ∇φ.
P is the canonical momentum, i.e., the momentum conjugate to the coordinate r and corresponding, in Quantum Mechanics, to the operator −i∇ (coordinate representation) The velocity, instead, is obtained from v= ∇ P H= 1 m
Thus, in the coordinate representation, the velocity components operators are v i = 1 m
, where the components ofAare not operators Thus, the desired commutators are
∂x j ψ( r ) = iq mc 2 ε i j k B k ψ( r ), whereε i j k is the Levi-Civita symbol.
Power-Law Potentials and Virial Theorem
A one-dimensional system is described by the Hamiltonian
Given an eigenstate|ψof this Hamiltonian, prove that
2V, whereT = p 2 /2meVis the potential energyV=λq n
Using the coordinate representation, it is easy to verify that q[p,H] =q[p,V] i λnq n =n i V.
So, ψ|V|ψ = − 1 i nψ|q[p,H]|ψ = − 1 i nψ|q pH−qHp|ψ = − 1 i nψ|q pH− [q,H]p−Hq p|ψ = 1 i nψ|[q,H]p|ψ.
Using the previous result, we obtain the desired relationship ψ|V|ψ = 1 i n i mψ|p 2 |ψ = 2 nψ|T|ψ.
Coulomb Potential and Virial Theorem
(a) Using the Schrửdinger equation, prove that, for every physical quantity of a quantum system, the Ehrenfest theorem holds: d dt = 1 ı[,H] +
To demonstrate the Virial theorem for the Coulomb potential, we apply the results derived from the operator rãp This theorem establishes a relationship between the expectation values of kinetic energy (T) and potential energy (V) in a stationary state.
(a) Call|ψ(t)the state vector of the physical system in the instantt The expectation value of is
= ψ(t)| |ψ(t), and, from the Schrửdinger equation i∂|ψ(t)
, where we have taken into account the fact that, in the Schrửdinger picture, time dependence of operators can only be explicit.
(c) As the system is in a stationary state andrãpdoes not depend on time, d rãp dt = ∂ rãp
Applying the Ehrenfest theorem, we obtain:
0= [rãp ,H] = [rãp ,T] + [rãp ,V] = [r ,T] ãp + rã [p ,V],
To calculate these two expectation values, we note that, from[r i ,p i ] =ı, we get
[r i ,p i 2 ] =2ıp i ⇒ [r ,T] ãp =2ıT, whereas we easily find that
By replacing these two relations in the previous one, the desired result is obtained.
Virial Theorem for a Generic Potential
(a) Using the Schrửdinger equation, prove that, for every quantity of a given physical system, the Ehrenfest theorem holds: d dt = 1 ı[,H] +
(b) Consider a system withN degrees of freedom and apply the previous result in the case of the operator
Q N i = 1 r i p i , with the purpose of demonstrating the Virial theorem relating the expectation values, in a stationary state, of the kinetic energyT and of the potential energy
(c) Apply the previous result to a one-dimensional harmonic oscillator.
(a) For the solution of this point, we refer to Problem1.10.
(b) Denoting the set of position coordinates withqand the set of conjugate momenta withp, we haveH(q,p)=T(p)+V(q), and
We calculate the commutators on the right side separately:
2m i ( p 2 i q i + 2i p i − p 2 i q i ) = i p i m i , while, from the power expansion in the variables q i of the potential V n c n q i n , we obtain
By replacing these two relations in Equation (1.1) we get
Applying the Ehrenfest theorem to the observableQ, under the hypothesis that the system is in a stationary state, we obtain dQ dt = 1 i[Q,H] +
Noting that neither Q, nor the probability distribution in a stationary state depends on time, we get
(c) The potential energy of a harmonic oscillator is
Then, the kinetic energy expectation value is given by
Feynman-Hellmann Theorem
Given a physical system, denote its Hamiltonian withHhaving eigenvaluesEand normalized eigenstates|Eso that
Assume that this Hamiltonian depends on a parameterλ,H =H(λ) As a conse- quence, its eigenvalues also depend onλ,E =E(λ).
Demonstrate that the following relationship holds:
AsEis the eigenvalue corresponding to|E, it results that
, as we wanted to prove.
Free Particles and Parity
For a one-dimensional free particle does the set of observables composed of the Hamiltonian and Parity constitute a complete set?
To every value of the energy of a free particle in a one-dimensional world, they correspond two linearly independent eigenstates that, in theX-representation, are given by the eigenfunctions: ψ p (x)= 1
Any linear superposition of eigenstates remains an eigenstate of the Hamiltonian, which is represented by the kinetic energy operator proportional to the second derivative and commutes with the Parity operator The linear combinations of ψₚ(x) and ψ₋ₚ(x) with definite parity yield cos(px/) with parity eigenvalue P = +1 and sin(px/) with parity P = -1 The eigenstates shared by the Hamiltonian and Parity eigenvalues are fully determined, establishing that these two operators constitute a complete set of commuting observables for the system.
Potential Step
Consider a particle of massmincident from the left on the potential (Fig.2.1)
V 0 ,forx>0. © Springer Nature Switzerland AG 2019
L Angelini, Solved Problems in Quantum Mechanics, UNITEXT for Physics,
Study its behavior for energy eigenvalues lesser and greater thanV 0 , determining the energy spectrum characteristics.
The energy eigenvalues (E) are always positive, as they must exceed the minimum potential To analyze this, we will solve the Schrödinger equation in the two regions where the potential remains constant.
In this region, the Schrửdinger equation can be written as:
The general solution for the wave function is expressed as ψ(x) = e^(i kx) + R e^(-i kx), where the coefficient of the incident plane wave traveling to the right is set to 1, and R represents the coefficient of the reflected wave This assumption allows us to derive the physical interpretation by calculating the probability density current from the relevant equation.
(e − ı kx +R ∗ e ı kx ) (ı ke ı kx −ı k Re − ı kx )
(e − ı kx +R ∗ e ı kx ) (e ı kx −Re − ı kx )
The probability density current consists of two time-constant components: the first represents a flow of 1 particle per second moving from left to right at speed v, while the second denotes a flow of |R|² particles per second traveling in the opposite direction.
In this region, we write the Schrửdinger equation as:
We distinguish two cases, depending on whether the energyE m is greater than or less thanV 0.
In this case,k is real Once again, the general solution is of the type ψ(x)=Te ı k x +Se − ı k x , (2.2) whereTandSare constants Since the particle comes from the left, it must beS =0.
In this analysis, we focus on a specific solution for particles traveling from left to right, represented by the term ψ(x)=Te ı k x This expression indicates a probability density current of |T|² particles per second moving through each point at speed v To find the coefficients R and T, we apply the continuity conditions for both the wave function and its first derivative at the boundary between the two regions, specifically at x=0, leading us to a linear system of equations.
Substituting RandT in the expression for the current, we obtain v|R| 2 =v k−k k+k
It is easy to verify that v(1− |R| 2 )=v |T| 2 , namely, the probability density current is the same in the two regions Note that:
• In the limit E0, that is, for k−→k , we find|R| 2 −→0 and|T| 2 −→1, confirming the intuition that, at high energy, the potential step can be neglected leading to the free particle motion.
• An important novelty emerges with respect to the classical behavior of the particles: the presence of a perturbation in the potential generates a finite probability of reflecting back the particle 1
We note that, being thatE−V 0 0 region.
In both cases,E >V 0 andEV 0 The spectrum is doubly degenerate In fact, even if we found only one solution to the case, this is the consequence of having placed the coefficient
S =0, in order to reproduce a physically reproducible situation There is, however, a solution that is linearly independent from this, corresponding to the sending of particles in the opposite direction to thexaxis.
In the scenario where E is less than V0, we set S equal to 0 to avoid a divergent solution at infinity This leads to one of the two linearly independent solutions of the eigenvalue equation falling outside the wave function space, indicating that the eigenvalues between 0 and V0 are non-degenerate.
Fig 2.2 Particle confined on a segment (infinite potential well)
A particle of massmconfined on a segment can be described by the potential (Fig.2.2)
+∞,elsewhere. also called the infinite potential well or one-dimensional box Find the energy spec- trum and the corresponding eigenfunctions.
The wave function is non-zero only within the range 0 < x < L, while in the region x > L, the potential becomes infinite, exceeding any fixed energy eigenvalue E To solve the Schrödinger equation in this region, we can assume a constant potential V₀ > E and then take the limit as V₀ approaches infinity This scenario mirrors the situation in problem (2.2), where E < V₀ in region II Consequently, the only solution for the wave function in this limit is ψ(x) = T e^(-χx).
A similar reasoning can be repeated in the xL and forx