Quantum Mechanics I Physics PDF Free Download

Quantum Mechanics I Physics 5701 Z. E. Meziani 02/24//2017 Physics 5701 Lecture Commutation of Observables and First Consequences of the Postulates

Outline 1 Commutation Relations 2 Uncertainty Relations 3 Time Evolution of the Mean Value of an Observable; Ehrenfest Theorem 4 Commutation of Observables 5 Application: A New Method to Solve the Harmonic Oscillator

Outline

Commutation Relations Commutation Relations In the general formalism of Hilbert space the commutation relations plays a very important role. We have to postulate the following fundamental relations. [ˆx, ˆp x] = i [ŷ, ˆp y] = i [ẑ, ˆp z] = i from these we can deduce many others. Consider the example of the orbital angular momentum ˆ L = ˆ r ˆ p, in the formalism of wave functions we have ˆL z = ( x i y y ), ˆLx = ( y x i z z ), ˆLy = ( z y i x x ) z We can establish from the above commutation relations the following relations [ˆLx, ˆL ] y = i ˆL z, [ˆLy, ˆL ] z = i ˆL x, [ˆLz, ˆL ] x = i ˆL y the results can be combined in a compact notation to give ˆ L ˆ L = i ˆ L

Commutation Relations Commutation Relations (continued) When we will evaluate the properties of angular momentum. We will take the above relation as the definition of the angular momentum. A first use of the commutation relations will lead to the proof of the uncertainty principle. More precisely to compute the uncertainty related to the prediction on the measurement of two observables that do not commute. We will also see how to get the eigenvalues and eigenvectors of the angular momentum. Example: Show that if then [ ] ˆx, Ĥ [ ] ˆp x, Ĥ Ĥ = ˆ p 2 2m + V (ˆ r ) = i ˆpx = i Ĥ m ˆp x = i x V (ˆ r )

Uncertainty Relations Uncertainty Relations A first use of the commutation relations will lead to the proof of the uncertainty principle. More precisely we want to compute the uncertainty related to the prediction on the measurement of two observables that do not commute. We need to prepare an ensemble of systems in the state ψ. We will thus perform the measurement of the A and B many times, however, each system will be measured one time.the set of results a and b will have a and b for mean values σ a and σ b for mean standard deviations. These are the numbers we want to calculate for a state ψ and two observables Â and ˆB.

Uncertainty Relations Uncertainty Relations (continued) Now consider for a state ψ the vector (Â + iλ ˆB ) ψ with λ real (Â + iλ ˆB ) ψ 2 = ψ (Â iλ ˆB )(Â + iλ ˆB ) ψ = ψ Â 2 ψ + λ 2 ψ B 2 ψ + iλ ψ [Â, ˆB ] ψ = σ 2 a + λ2 σ 2 b + iλ ψ [Â, ˆB ] ψ Â and ˆB being Hermitian, their commutator is anti-hermitian thus, the last term is real. The above expression being the norm of a vector, it has to be positive for any λ. The discriminant in λ of the trinomial has to be negative (Schwarz inequality). Here we also use the fat that [Â, ˆB ] = [Â, ˆB]. σ a σ b = a b 1 ] ψ [Â, ˆB ψ 2 This is the general form of the Heisenberg uncertainty relation.

Time Evolution of the Mean Value of an Observable; Ehrenfest Theorem Time Evolution of the Mean Value of an Observable; Ehrenfest Theorem Here we will calculate the time evolution of the mean value for a quantity. We will use the results on the conjugate variable q i and p i and find a form similar to the Hamilton equations of classical mechanics. We will thus be able to make the connection between classical and Quantum mechanics We have already seen that the mean value of measurements performed on the observable Â of a system prepared on ψ is a ψ = ψ Â ψ Now lets take the time derivative of this quantity ( ) d d dt a = dt ψ Â ψ + ψ Â ψ + ψ Â t ( ) d dt ψ

Time Evolution of the Mean Value of an Observable; Ehrenfest Theorem Time Evolution of the Mean Value of an Observable (continued) i) If Â and Ĥ commute ψ the mean value a is time independent, it is a constant of motion ii) If ψ is eingenstate of Ĥ, for any Â, a is time independent, ψ is thus a stationary state iii) Case of a particle in a scalar potential V ( r): q i (i = 1, 2, 3) are x, y, z coordinate in space of the position r p i (i = 1, 2, 3) are p x, p y, p z coordinate of the momentum p The operators ˆq i and ˆp i fulfill the following commutation relations [ˆq i, ˆq j ] = 0 [ˆp i, ˆp j ] = 0 [ˆq i, ˆp j ] = i δ ij Now Ĥ is a function of the operators ˆq i and ˆp i, then from the above relations it is easy to establish that [ˆq i, ˆp m i ] = m(i )ˆpm 1 i [ˆp i, ˆq n i ] = n(i )ˆqn 1 i

Time Evolution of the Mean Value of an Observable; Ehrenfest Theorem Time Evolution of the Mean Value of an Observable (continued) Note that the above equations are the Hamilton equations where all classical quantities are replaced by the mean of the corresponding observables in the state ψ In classical mechanics, starting with a Hamiltonian we have H p i = p i m = dq dt H q i = V ( r) q i H = p2 2m + V ( r) = dp i dt definitionof v definitionof F = m γ If it was possible to reduce indefinitely the dimensions of the wave packet in q i and p i, then Ehrenfest s equations would give Hamilton s equations. Comment. Classical limit: For a macroscopic system, it is necessary that the results of quantum mechanics join those of classical mechanics.the Ehrenfest theorem garanties this correspondence. In fact, for a particle, if 0 it is possible to reduce the wave packet ψ in position and momentum spaces x p x 0 Ehrenfest s eq. = Hamilton s eq. This can be considered as a "justification" of the "principle of correspondence" which guides us in the choice of quantum observables for the physical quantities that have an classical analogue.

Commutation of Observables Commutations of Observables Theorem: If two observables Â and ˆB commute, there exist in E a basis composed of eigenvectors common to Â and ˆB and reciprocally.

Commutation of Observables Commutations of Observables Theorem: If two observables Â and ˆB commute, there exist in E a basis composed of eigenvectors common to Â and ˆB and reciprocally. { compatible Â, Observables: ˆB commute incompatible Â, ˆB do not commute Example: Consider an atomic system and the two observables ˆL 2 and ˆL z associated with L 2 (square of the angular momentum) and L z (projection of the angular momentum on Oz). For simplicity assume that the Hamiltonian is time independent and that it commute with ˆL 2 and ˆL z One can prove that ˆL 2 and ˆL z commute. This means there is a basis { i, l, m } such that each basis vector is eigenvector of ˆL 2 and ˆL z. We will see that ˆL 2 i, l, m = l(l + 1) 2 i, l, m and ˆLz i, l, m = m i, l, m Now we will comment on the postulate of the measurement in the case where in the initial state of the atomic system ψ is any state ψ = ilm c ilm i, l, m

Commutation of Observables Commutations of Observables (continued)

Commutation of Observables Commutations of Observables (continued) ii) Now we perform the measurement on ˆL z. Again one cannot predict the result but the probability P (m) is different from the initial one P(m). The measurement gives m the state after measurement is ψ = c i c il m i, l m iii) From this moment any measurement of ˆL 2 and ˆL z will give the same l (l + 1) and m. This is the fundamental consequence of the commutation of the two operators ˆL 2 and L z. Case of many observables Â, ˆB and Ĉ that commute: The generalization of the theorem is immediate. We said that Ĥ commute with L2 and L z, we could have chosen a basis where the vectors of the basis are also eigenvectors of Ĥ. If i, n, l, m is such a basis we could have measured the energy, L 2 and L z. Repeating the measurement will give always the same results. Complete Set of Commuting Observables (CSCO): A quantum system is "perfectly" defined if one can the measurement of all quantities of a CSCO. To a phase factor its state ψ is thus defined. Nevertheless, the measurement of a physical quantity whose observable does not commute with the operators of the CSO cannot be predicted. We will talk about "totally prepared quantum system". Example: The state nlm of the hydrogen atom is determined if we neglect the spin of the electron and proton as we will see later.

Application: A New Method to Solve the Harmonic Oscillator Application: A New Method to Solve the Harmonic Oscillator We will illustrate the functioning of the postulate and the method is based on commutator algebra. Consider the Hamiltonian Ĥ = ˆp2 2m + 1 2 mω2 ˆx 2 we make the following change of variables we obtain with ˆX = mω ˆx ˆP = 1 mω ˆp Ĥ = ωĥ Ĥ = 1 ( 2) ˆX2 + ˆP 2 we need to solve the eingenvalue problem Ĥ ν = ɛ ν ν we have assumed the ɛ ν eigenvalues non-degenerate (this can be shown) in other words Ĥ or Ĥ constitute a CSCO by itself. From the commutation relations of ˆX and ˆP we get [ ˆX, ˆP ] = i

Application: A New Method to Solve the Harmonic Oscillator A New Method to Solve the Harmonic Oscillator (continued) a) Operators â, â and ˆN

Application: A New Method to Solve the Harmonic Oscillator A New Method to Solve the Harmonic Oscillator (continued) a) Operators â, â and ˆN the solution of the problem is made easy by the introduction of the following operators â = 1 2 ( ˆX + i ˆP ) â = 1 2 ( ˆX i ˆP ) first, we note that [ â, â ] = 1 consider the operator ˆN = â â = 1 2 ( ˆX2 + ˆP 2 1) which satisfies the following commutation relations [ ] [ ] ˆN, â = â ˆN, â = â

Application: A New Method to Solve the Harmonic Oscillator A New Method to Solve the Harmonic Oscillator (continued) b) Determination of the eigenvalues Assume ν are the eigenvalues of ˆN corresponding to eigenvectors ν, we go back to the initial Hamiltonian writing Ĥ ν = ( ν + 1 2 ) ω ν i) the eigenvalues of ˆN are positive or zero â ν 2 = ν â â ν = ν ˆN ν = ν ν ν = ν ( ν ) 2 it is obvious that ν = 0 then â ν = 0 since the norm of a vector is zero only if the vector is equal to zero. ii) â ν is eigenvector of ˆN with eigenvalue ν 1 Using the commutation relation of ˆN with â and â we have ˆN(â ν ) = â ˆN ν â ν = νâ ν â ν = (ν 1)â ν iii) The same reasoning leads to show â is eigenvector of ˆN with ν + 1 being the associated eigenvalue and â ν 2 = (ν + 1) ν 2 Assume a vector ν which corresponds to the eigenvalue ν 0, we are sure we will find an integer n such that n 1 < ν n if we act on ν with the operator (â) n we will find an eigenvector which correspond to the eigenvalue ν = ν n but from i) ν = ν n 0, thus ν n. Therefore, ν = n, the eigenvalues of ˆN are integers and non negative. We thus find the known energy levels of the harmonic oscillator.

Application: A New Method to Solve the Harmonic Oscillator A New Method to Solve the Harmonic Oscillator (continued) c) Determination of the eigenstates

Application: A New Method to Solve the Harmonic Oscillator A New Method to Solve the Harmonic Oscillator (continued) c) Determination of the eigenstates i) The ground state 0, satisfies the equation 1 [ ] â 0 = 0 thus ˆX + i ˆP 0 = 0 2 In the x representation this relation becomes ( mω + d ) φ 0(x) = 0 dx where φ(x) = φ 0 This equation can be solved immediately giving φ 0(x) = c 0e mωx2 2 where c 0 is a normalization constant chosen such that φ 0φ0dx = 1. ii) Excited states If we assume that all states n are normalized to n n = 1, we can show using earlier findings that â n = n n 1 â n = n + 1 n + 1 (1) leading to the name of "annihilation" operator for â and "creation" operator for a since they make the transition from a state of energy (n + 1/2) to (n + 1/2 ± 1) and annihilate or create a quantum fo energy ω. Similarly the operator ˆNcorresponds to the number of quantas of the state n.

Application: A New Method to Solve the Harmonic Oscillator A New Method to Solve the Harmonic Oscillator (continued) We see that the series of states n is generated from the ground state 0 through the repeated action of the operator â. n 1 n! (â ) n 0 When expressed in the { x } representation allows us to find the wave function φ n(x) of the energy state (n + 1/2) ω starting from the ground state: [ φ n(x) = 1 ] n 1 mω x n! 2 n d φ 0 (x) mω dx We obtain thus a compact formula for the Hermite polynomials.