� � 8.04 Quantum Physics Lecture XIX We then have ˆH|˜1� = ˆHˆa†|0� = � � ˆH, ˆa† � + ˆa† ˆH � |0� = � ¯hωˆa† + ˆa† 1 2 ¯hω0 � |0� = 3 2 ¯hωˆa†|0� = 3 2 ¯hω|˜1�, (19-1) (19-2) (19-3) (19-4) (19-5) 3 1 = ˆ 0 is also an energy eigenstate, but with eigenenergy � � 2 ˜ ˜ † Similarly, we can show that 2 = ˆ 1 is also an energy eigenstate, but with � � a Consequently, we can construct a ladder of (yet to be normalized) |˜1 a† i.e.,¯hω instead of¯hω|0�.2 for|energy ||5 2 ¯hω etc. energy eigenstates n˜� by ||n˜� = aˆ† n |0� (19-6) with �� 1 En = n + hω. ¯(19-7) 2 aˆ(ˆa†) is called the lowering (raising) operator, it lowers (raises) the energy by ¯hω. Figure I: ˆa,ˆa† are sometimes called “ladder operators” since they take us up and down the ladder of energy eigenstates. When describing a monochromatic electromagnetic field quantum mechanically, we can associate the frequency ω with a harmonic oscillator of that frequency. For non-interacting particles (such as photons) a state with n photons can be associated with the n-th eigenstate of the HO with n. The ground state then corresponds to an Massachusetts Institute of Technology XIX-1 8.04 Quantum Physics Lecture XIXempty mode (no photons, n = 0), however there is still a finite energy1 2 ¯hω that we associate with vacuum fluctuations of the electromagnetic field. In this context, aˆ† and ˆa are called creation and annihilation operators, respectively, since they create and annihilate photons, or more generally, arbitrary non-interacting bosonic particles. Normalization of HO energy eigenstates Let us assume that the ground state |0� is already chosen to be properly normalized: ∗dxu0�0|0� = 1. � Note. Remember that �0|0� denotes �0|0� =(x)u0(x). How long is the state |1˜� =ˆa†|0�? �1|1� = �aˆ†0|aˆ†0� (19-8) = �0|aˆ|aˆ†0� (19-9) = �0|a� ˆaˆ†|0�� (19-10) = �0| ˆa† a†ˆ|0� �� (19-11) a, ˆ+ˆa= �0|1+ˆa†aˆ|0�→ aˆ|0� = 0 (19-12) = 1 (19-13) The state |1˜� is already normalized, so we can write: |1� =ˆa†|0�→ normalized eigenstate (19-14) What about |2˜� =ˆa†|1˜� =ˆa†|1�? �˜2|2˜� = �aˆ†1|aˆ†1� (19-15) = �1|� aˆaˆ†|1� � (19-16) = �1| aˆ†aˆ+1 |1� �� (19-17) = �1|aˆ†|0� +1 → aˆ|1� = |0� (19-18) = �1|1� + 1 (19-19) = 2 (19-20) Massachusetts Institute of Technology XIX-2 � � � �� � � � 8.04 Quantum Physics Lecture XIXThen the properly normalized second excited state is 11 ��2 |2� = √2|2˜� = √2 aˆ† |0�. (19-21) We can show, in general, (see PS) that the length squared of the state |n˜� = aˆ† n |0�is �n˜|n˜� = n!. Consequently, the n-th normalized eigenstate is 1 ��n |n� := √n! aˆ† |0� . (19-22) We can also show (see PS) that aˆ|n� = √n|n − 1�, (19-23) aˆ†|√n +1|n +1�. (19-24) n� = From operators back to spatial wavefunctions The condition on the ground state |0�,ˆa|0� = 0, reads in position space using our definition of the annihilation operator, xˆpˆmω i aˆ= + i = xˆ+ p,ˆ(19-25) x0 p0 2¯h √2¯hmω mω ih¯∂ ˆ2¯hx + √2¯hmω i∂x (19-26) au0(x)= u0(x)=0 ∂ mωx +¯hu0(x)=0. (19-27) ∂x mω 2 2¯hThe simple DE has the solution u0(x)= ce− xwith normalization 1 = c2 π¯h .mω Consequently, the normalized ground-state wavefunction is ��1 mω 2mω 4 2¯u0(x)= e− h x. (19-28) πh¯The normalized n-th eigenstate can be obtained from 1 ��|n� = √n! aˆ† n |0� (19-29) or �� �1 mω ih¯∂ n un(x)= √n! 2¯hx −√2¯hmωi∂x u0(x). (19-30) Massachusetts Institute of Technology XIX-3 � � � � � � � � � � � � � � � � � � � � � � � � � 8.04 Quantum Physics Lecture XIXCommutators, Heisenberg uncertainty, and simultanneou eigenfunctions The fact that ˆp = ¯h∂ in the position representation (or ˆx = ih¯∂ in the momentum i ∂x ∂p representation) implies pˆˆp xψ(x)= xˆˆ,xψ(x)= ˆˆpψ(x)= x pψ(x) (19-31) i.e., ˆx and ˆp do not commute. Define the difference between ˆpxˆand ˆxpˆas the commuttato p,ˆxˆ= pˆxˆ− xˆp.ˆ(19-32) Here: �� h¯p,ˆxˆ= (c-number) (19-33) i → In general, ˆB =ˆB − BˆAˆis an operator. The commutator is linear. A, ˆA ˆc1Aˆ1 + c2Aˆ2,Bˆ= c1 Aˆ1,Bˆ+ c2 Aˆ2,Bˆ(19-34) Other useful relations ˆˆB, Aˆ= − A,Bˆ(19-35) AˆˆC =ˆˆC +ˆCBB, ˆA B, ˆA, ˆˆ(19-36) Simultaneous eigenfunctions Consider a free particle. The plane waves ψ(x)= e±ikx are simultaneous eigenfunc¯tions of energy with eigenvalue h2k2 ,2m h¯2 ∂2 h¯2k2 Heˆ±ikx = −2m ∂x2 e±ikx =2me±ikx , (19-37) and of momentum with eigenvalue ±¯hk, pe±ikx h¯∂e±ikx hke±ikx ˆ= = ±¯. (19-38) i ∂x Note. If we had chosen cos(kx), sin(kx), these would have also been energy eigenh22k¯functions with eigenvalue 2m , but not momentum eigenfunctions. Massachusetts Institute of Technology XIX-4 8.04 Quantum Physics Lecture XIXHowever, since cos(kx) and sin(kx) are degenerate (i.e., have the same energy eigenvalue), it is possible to choose linear combinations of degenerate eigenstates e±ikx = cos(kx) ± i sin(kx) that are simultaneous eigenstates of momentum. In the potential well, on the other hand, the energy eigenstates were not simultaneous eigenstaate of momentum. In general, we have: Theorem 19.1. Two Hermitian operators Aˆ, Bˆhave a set of simultaneous eigenfuncction if and only if they commute. Proof. ” ” Assume a complete set {uab} of simultaneous eigenfunctions is found, ⇒i.e., ˆAuab = auab (19-39) Buˆab = auab (19-40) a,b, eigenvalues. Then [ A,ˆBˆ]uab =(ab − ba)uab = 0 for all eigenfunctions → A, ˆ[ˆB]= 0. ” ” See Gasiorowicz, 5-4.⇐Since only an eigenstate of Aˆwill have a definite outcome when a measurement of Aˆis made, this means that ΔA and ΔB can always be simultaneously made zero only when Aˆand Bˆcommute. Theorem 19.2. One can prove that in any chosen state ψ, (ΔA)2 ψ(ΔB)2 A,Bˆ]�ψ 2 (19-41) ψ ≥�i[ˆfor any two Hermitian operators Aˆ, Bˆ. Proof. see Gasiorowicz, online supplement SA. For ˆx,ˆp, we have (Δx)2 ψ(Δp)2 ψ ≥ 41 �iih¯�2 ψ = h¯4 2 , (19-42) where the RHS does not depend on the state ψ. This is another derivation of the ¯Heisenberg uncertainty relation ΔxΔp ≥ h .2 The Schr¨odinger equation in three dimensions ˆHψ(r)= Eψ(r) → SE in 3D (19-43) with pˆ2 = pˆ2 +ˆp 2 +ˆp 2 (19-44) xyz Massachusetts Institute of Technology XIX-5 � � � � � � � � � � � � � � � � 8.04 Quantum Physics Lecture XIXh¯∂h¯∂h¯∂ pˆ= ,, in the position representation (19-45) i∂xi∂y i∂z → The SE then reads 2 = ∂2 + ∂2 + ∂2 ,∂x2 ∂y2 ∂z2 h¯22−2m�+ V (r) ψ(r)= Wψ(r) → SE in 3D (19-46) Spherically symmetric potential If the potential is spherically symmetric, V (r)= V (r), then it is convenient to work in spherical coordinates, where we can write ∂2 2 ∂ 1 ∂2 ∂ 1 ∂2 2�= ∂r2 + r ∂r + r2 ∂θ2 + cot θ∂θ + sin2 θ ∂φ2 (19-47) We define an operator via ∂2 ∂ 1 ∂2 Lˆ2 = −h¯2 ∂θ2 + cot θ∂θ + sin2 θ ∂φ2 . (19-48) Lˆwill be the operator associated with angular momentum. ∂2 2 ∂ Lˆ2 2�= ∂r2 + r ∂r − h¯2 r(19-49) 2 Since V (r) does not depend on θ, φ, we try an ansatz. ψ(r)= R(r)Y (θ,φ) (19-50) Then, � ��� �� h¯22 h¯2 ∂2 2 ∂ −2m�+ V (r) ψ(r)= −2m ∂r2 r ∂r + V (r) R(r)Y (θ,φ) (19-51) L2 + R(r)Y (θ,φ) (19-52) 2mr2 = ER(r)Y (θ,φ) (19-53) As before, when deriving the time-indepenedent SE, we divide by R(r)Y (θ,φ) = 0. h¯2 ∂2 2 ∂ 1 L2 ··· = −2m ∂r2 + r ∂r + V (r) R(r)+ Y (θ,φ)2mr2 R(r)Y (θ,φ) (19-54) = E (19-55) Massachusetts Institute of Technology XIX-6 � � � � � � � � 8.04 Quantum Physics Lecture XIXThe LHS can only be a constant for all θ, φ if the second term does not depend on θ, φ. We arrive at two equations: Lˆ2 const Y (θ,φ)= Y (θ,φ)= EL(r)Y (θ,φ) (19-56) 2mr2 2mr2 1 h¯2 ∂2 2 ∂ const R(r) −2m ∂r2 + r ∂r + V (r) R(r)+ 2mr2 = E (19-57) h¯2 ∂2 2 ∂ const −2m ∂r2 + r ∂r + V (r)+ 2mr2 R(r)= ER(r) (19-58) where EL = const is the energy associated with the angular dependence of the wave2mmr function. Massachusetts Institute of Technology XIX-7