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Class 2: Operators, Stationary States

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Class 2: Operators, Stationary States Class 2: Operators, stationary states Operators In quantum mechanics much use is made of the concept of operators. The operator associated with position is the position vector r. As shown earlier the momentum operator is −iℏ ∇ . The operators operate on the wave function. The kinetic energy operator, T, is related to the momentum operator in the same way that kinetic energy and momentum are related in classical physics: p⋅ p (−iℏ ∇⋅−) ( i ℏ ∇ ) −ℏ2 ∇ 2 T = = = . (2.1) 2m 2 m 2 m The potential energy operator is V(r). In many classical systems, the Hamiltonian of the system is equal to the mechanical energy of the system. In the quantum analog of such systems, the mechanical energy operator is called the Hamiltonian operator, H, and for a single particle ℏ2 HTV= + =− ∇+2 V . (2.2) 2m The Schrödinger equation is then compactly written as ∂Ψ iℏ = H Ψ , (2.3) ∂t and has the formal solution iHt Ψ()r,t = exp − Ψ () r ,0. (2.4) ℏ Note that the order of performing operations is important. For example, consider the operators p⋅ r and r⋅ p . Applying the first to a wave function, we have (pr⋅) Ψ=−∇⋅iℏ( r Ψ=−) i ℏ( ∇⋅ rr) Ψ− i ℏ( ⋅∇Ψ=−) 3 i ℏ Ψ+( rp ⋅) Ψ (2.5) We see that the operators are not the same. Because the result depends on the order of performing the operations, the operator r and p are said to not commute. In general, the expectation value of an operator Q is Q=∫ Ψ∗ (r, tQ) Ψ ( r , td) 3 r . (2.6) V 1 If Q is associated with an observable quantity, then its expectation value must be a real number, i.e. ∗ Q= Q . We can check to see if this is the case for the momentum operator for which p=∫ Ψ∗ ( −ih ∇Ψ) d 3 r . (2.7) V Now p∗ =Ψ∫()ih ∇Ψ∗ dih3 rr = ∫ Ψ∇Ψ ∗ dih 3 = ∫ ∇ΨΨ( ∗∗) −Ψ∇Ψ d 3 r V V V (2.8) =ih∫ ∇ΨΨ()∗ dih3r − ∫ Ψ∇Ψ ∗ d 3 r . V V ∗ By applying the divergence theorem to the first integral we see that is zero. Hence p= p , and we conclude that the momentum is an observable quantity. Uncertainty In a statistical sense, the degree of uncertainty in a measured quantity is often given by the standard deviation, σ . The standard deviation is the square root of the variance. Thus the uncertainty (standard deviation) in the x component of the particle’s position is given by 2 2 2 σ x =x − x . (2.9) Similarly, the uncertainty in the x component of the particle’s momentum is given by 2 2 2 σ p=p x − p x . (2.10) Because the operators x and px do not commute the product of the uncertainties in the measurements of the x-components of position and momentum must be greater than or equal to ℏ 2. However there is no such restriction on the uncertainty in orthogonal components of position and momentum, i.e. ℏ σ σ ≥ , (2.11) x p 2 but σy σ p ≥ 0. (2.12) Equation (2.11) is an example of Heisenberg’s Uncertainty Principle (HUP). The existence of the HUP can be motivated by a simple ‘thought’ experiment. Consider a particle travelling in 1 dimension. Suppose we want to determine its position and momentum at some instant by shining low intensity light of wavelength λ on it. The light is assumed to create an image of the particle on a screen. By measuring 2 the position of the image at slightly different times, the momentum of the particle can be measured. However, a scattering of a single photon will change the particle’s momentum by an amount h ∆p ∼ , (2.13) λ and because of diffraction effects the particle’s image cannot be smaller than the wavelength λ. Thus the uncertainty in position is ∆x ∼ λ. (2.14) Combining the last two expressions, we see that ∆p ∆ x∼ h . The precise position and momentum of the particle cannot be measured simultaneously. Stationary states The Schrödinger equation for a single particle ∂Ψ ℏ2 iℏ =− ∇Ψ+2 V ()r Ψ . (2.15) ∂t2 m has a number of properties that allow it to be solved in relatively straightforward ways. It is a linear equation which means that any linear combination of solutions is a solution. Also if the potential energy is independent of time, as we have assumed, solutions can be found by the method of separation of variables. For simplicity, we begin by considering one dimensional problems for which the Schrödinger equation is ∂Ψℏ2 ∂ 2 Ψ iℏ =− + V() x Ψ . (2.16) ∂t2 m ∂ x 2 Separation of variables The method of separation of variables is to look for solutions of form Ψ( xt,) = ψ ( xTt) ( ) . (2.17) Substituting this into equation (2.16), we find dTℏ2 d 2 ψ iℏψ= − T + VT ψ . (2.18) dt2 m dx 2 The next step is to divide by ψT, which gives 3 1dTℏ2 1 d 2 ψ iℏ = − + V . (2.19) T dt2 mψ dx 2 The left hand side is a function of t alone and the right hand side is a function of x alone. The only way that this is possible is for both sides to equal the same constant. (To see why choose a particular value of x, for example, to evaluate the right hand side. The left hand side must equal this value for all values of t, so that the left hand side is equal to a constant. The right hand side is then equal to the same constant.) Denote the constant by E. Then dT iE = − T, (2.20) dt ℏ which has solution iE T() t=exp − t . (2.21) ℏ The function ψ ( x) is a solution of the time independent Schrödinger equation ℏ2d 2 ψ − +Vψ = E ψ . (2.22) 2m dx 2 This equation cannot be solved unless the potential energy function V( x ) and boundary conditions are specified. As we will see later, a common situation is that physical solutions exist only for a discrete set of values of the constant E. Suppose we have found the solution ψ ( x) of equation (2.22) corresponding to a particular value of E. The separable solution of equation (2.16) is then iE Ψ()()xt, =ψ x exp − t . (2.23) ℏ Since the probability density Ψ∗ ( xt,) Ψ( xt , ) =∗ψ( x) ψ ( x ) (2.24) is time independent, the separable solution corresponds to a situation in which nothing changes with time. The separable solutions correspond to stationary states . Also, for a stationary state, the expectation value of the Hamiltonian is equal to the constant E. Hence E is the mechanical energy of the particle in a stationary state. The complete solution of the Schrödinger equation The complete solution of the Schrödinger equation is a linear combination of separable solutions, 4 ∞ iE n Ψ()()xt, =∑ cxnψ n exp − t , (2.25) n=1 ℏ ⋯ where E1, E2, E3, … are the values of E for which separable solutions exist, and ψ1, ψ 2 , ψ 3 , are the corresponding solutions of equation (2.22). The cn are complex constants, which can be determined from initial conditions. Note that the general solution does not correspond to a stationary state, as can be seen by considering a solution which is the sum of two separable solutions, iE1 iE 2 Ψ=()()xtcx,11ψ exp −+ tcx 22 ψ () exp − t . (2.26) ℏ ℏ The probability density is ∗ ∗∗∗∗iEt1ℏ iEt 2 ℏ − iEt 1 ℏ − iEt 2 ℏ ΨΨ=( xt,) ( xtce , ) ( 11ψ + c 22 ψ e)( ce 11 ψ + ce 22 ψ ) (2.27) 2 2 2 2 ∗∗iEEt()1− 2 ℏ ∗∗ − iEEt()1 − 2 ℏ =++c11ψ c 2 ψ 2 cce 1212 ψψ + cc 1212 ψψ e . Note that because the last term is the complex conjugate of the third term, the probability density is a real function, but provided E1≠ E 2 , it is not time independent. Problem 2.1 Prove the following three theorems: (a) For normalizable solutions, the separation constant E must be real. (b) The time-independent wave function ψ ( x) can always be taken to be real. (c) If V( x ) is an even function, then ψ ( x) can always be taken to be either even or odd. (a) Suppose the separation constant has an imaginary part such that E= E0 + i Γ . The separable solution is then i( E0 + i Γ ) iE Γ Ψ=()()xtx,ψ exp − tx = ψ () exp − 0 tt exp . (2.28) ℏ ℏ ℏ Hence the probability density changes with time, and an initially normalized solution will not remain normalized. (b) Suppose, for a given E, the solution ψ has an imaginary part. Since the Hamiltonian is a real operator, ψ ∗ is also a solution of the time independent Schrödinger equation for the same value of E. Because the time independent Schrödinger equation is linear, the real function ψ+ ψ ∗ is also a solution, as is the real function i (ψ− ψ ∗ ). 5 (c) For V( x ) an even function, if ψ ( x) is a solution of the time independent Schrödinger equation, then so will be ψ (−x). The even function ψ( x) + ψ ( − x ) and the odd function ψ( x) − ψ ( − x ) are also solutions. Problem 2.2 Show that E must exceed the minimum value of V( x ) for every normalizable solution of the time independent Schrödinger equation. From part (b) of the previous problem, we can take ψ to be real. By writing the time independent Schrödinger equation as d2ψ 2 m =()V − E ψ , (2.29) dx 2ℏ 2 we see that if E< V everywhere, then ψ and it second derivative always have the same sign.
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