Chapter 2.Some simple cases1. Translational motion (free particles)2. Particle in a box3.Theharmonicoscillator
Chapter 2. Some simple cases 1. Translational motion (free particles) 2. Particle in a box 3. The harmonic oscillator
2.1FreeparticlesClassically, a particle moving in one dimension without external forces has constantmomentum.QuantummechanicallyweexpecttobeabletofindstatesofdefinitemomentumIf the motion is along the x axis we need to look for solutions of the eigenvalue equationp.y (x)= py/ (x)That is,dit-y=pydxThesolutiontothis equationis,=exp(ipx / h)Remember that p is an eigenvalue - a constant with (in this case) dimensions of momentum
2.1 Free particles Classically, a particle moving in one dimension without external forces has constant momentum. Quantum mechanically we expect to be able to find states of definite momentum. If the motion is along the x axis we need to look for solutions of the eigenvalue equation That is, The solution to this equation is Remember that p is an eigenvalue — a constant with (in this case) dimensions of momentum. p x p x ˆ x d d i p x p =exp / ipx
TheHamiltonian for a free particle contains only the kinetic energy term:H==-_ d?2m dx?2mThe wavefunction , is an eigenfunction of this operator too:? d?h?ipx/hjpx/hHy, 2m dx?2m21Its energy is p? /2m, just as we would expect for a particle with momentum p.Notice that a particle with momentum -p i.e. with wavefunctionΦ-p = exp(-ipx/h) — has the same energy p2 /2m
The Hamiltonian for a free particle contains only the kinetic energy term: The wavefunction ψp is an eigenfunction of this operator too: Its energy is 𝑝 2 /2𝑚, just as we would expect for a particle with momentum p. Notice that a particle with momentum −p — i.e. with wavefunction 𝜓−𝑝 = exp(−𝑖𝑝𝑥/ℏ) — has the same energy 𝑝 2 /2𝑚. 2 2 2 2 ˆ d ˆ 2 2 d p H T m m x 2 2 2 2 2 / / 2 d e e 2 d 2 2 ipx ipx p p ip p H m x m m
Thetime-independent Schrodingerequationist? d'yHy=Ey2m dx?andthegeneralsolutionofthisisy = aeikr + be-ikrwhere k = V2mE/h.If E = p?/2m, then k = p/h and we arrive at the resulty = ay,+by-pThis superposition or linear combination of two wavefunctions, both with energy p?/2m.is also an eigenfunction of H with energy p?/2m, for any values of the constants a and bHowever it is not an eigenfunction of the operator px, unless a = O or b = O, so it doesn't have adefinitemomentum. We can write the same wavefunction in the formy = Asin(px/h)+ Bcos(px/h)
The time-independent Schrödinger equation is and the general solution of this is where 𝑘 = 2𝑚𝐸/ℏ. If 𝐸 = 𝑝 2 /2𝑚, then 𝑘 = 𝑝/ℏ and we arrive at the result This superposition or linear combination of two wavefunctions, both with energy 𝒑 𝟐 /𝟐𝒎, is also an eigenfunction of H with energy 𝒑 𝟐 /𝟐𝒎, for any values of the constants a and b. However it is not an eigenfunction of the operator px , unless a = 0 or b = 0, so it doesn’t have a definite momentum. We can write the same wavefunction in the form 2 2 2 d 2 d H E m x e e ikx ikx a b p p a b A px B px sin / cos /
2.2 Particle in a boxConsider a particle in a box': suppose that the potential is zero for O < x < a and infiniteoutsidethisrange.TheSchrodingerequationish?dHy=(↑+V)y=+VW=Ey2m dx2Outside the box, where Vis infinite, the only solution is y = O.Inside the box, possiblesolutions are exp(ipx/h) and exp(-ipx/h) both with energy E = p?/2m.However the wavefunction has to be continuous, so it must be zero at both ends of the boxWe can achieve this by using the wavefunction Asin(px/h) + Bcos(px/h)If the wavefunction is to be zero when x = O, then B = 0. If it is to be zero when x = a, thenpa=n元for integer nh
2.2 Particle in a box Consider a ‘particle in a box’: suppose that the potential is zero for 0 < x < a and infinite outside this range. The Schrödinger equation is Outside the box, where V is infinite, the only solution is ψ = 0. Inside the box, possible solutions are exp(𝑖𝑝𝑥/ℏ) and exp(−𝑖𝑝𝑥/ℏ) both with energy 𝐸 = 𝑝 2 /2𝑚. However the wavefunction has to be continuous, so it must be zero at both ends of the box. We can achieve this by using the wavefunction 𝐴𝑠𝑖𝑛(𝑝𝑥/ℏ) + 𝐵𝑐𝑜𝑠(𝑝𝑥/ℏ). If the wavefunction is to be zero when x = 0, then B = 0. If it is to be zero when x = a, then 2 2 2 d ˆ ˆ 2 d H T V V E m x pa n for integer n