# 1 Lecture 3: Operators in Quantum Mechanics

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1 1 Lecture 3: Operators in Quantum Mechanics 1.1 Basic notions of operator algebra. In the previous lectures we have met operators: ˆx and ˆp = i h they are called fundamental operators. Many operators are constructed from ˆx and ˆp; for example the Hamiltonian for a single particle: Ĥ = ˆp2 2m + ˆV (ˆx) where ˆp 2 /2m is the K.E. operator and ˆV is the P.E. operator. This example shows that we can add operators to get a new operator. So one may ask what other algebraic operations one can carry out with operators? The product of two operators is defined by operating with them on a function. Let the operators be Â and ˆB, and let us operate on a function f(x) (one-dimensional for simplicity of notation). Then the expression Â ˆBf(x) is a new function. We can therefore say, by the definition of operators, that Â ˆB is an operator which we can denote by Ĉ:Ĉ is the product of operators Â and ˆB. The meaning of Â ˆBf(x) should be that ˆB is first operating on f(x), giving a new function, and then Â is operating on that new function. Example: Â = ˆx and ˆB = ˆp = i hd/dx, then we have Â ˆBf(x) = ˆxˆpf(x) We can of course also construct another new operator: Then, by definition of the operator product, ˆpˆx ˆpˆxf(x) means that ˆx is first operating on f(x) and then ˆp is operating on the function ˆxf(x). Compare the results of operating with the products ˆpˆx and ˆxˆp on f(x): ( (ˆxˆp ˆpˆx)f(x) = i h x df(x) dx and hence by the product rule of differentiation: (ˆxˆp ˆpˆx)f(x) = i hf(x) d ) dx (xf(x)) 1

2 and since this must hold for any differentiable function f(x), we can write this as an operator equation: ˆxˆp ˆpˆx = i h Thus we have shown that the operator product of ˆx and ˆp is non-commuting. Because combinations of operators of the form Â ˆB ˆBÂ do frequently arise in QM calculations, it is customary to use a short-hand notation: [Â, ˆB] Â ˆB ˆBÂ and this is called the commutator of Â and ˆB (in that order!). If [Â, ˆB] 0, then one says that Â and ˆB do not commute, if [Â, ˆB] = 0, then Â and ˆB are said to commute with each other. An operator equation of the form of [Â, ˆB] = something is called a commutation relation. is the fundamental commutation relation. [ˆx, ˆp] = i h 1.2 Eigenfunctions and eigenvalues of operators. We have repeatedly said that an operator is defined to be a mathematical symbol that applied to a function gives a new function. Thus if we have a function f(x) and an operator Â, then Âf(x) is a some new function, say φ(x). Exceptionally the function f(x) may be such that φ(x) is proportional to f(x); then we have Âf(x) = af(x) where a is some constant of proportionality. In this case f(x) is called an eigenfunction of Â and a the corresponding eigenvalue. Example: Consider the function f(x, t) = e i(kx ωt). This represents a wave travelling in x direction. Operate on f(x) with the momentum operator: ˆpf(x) = i h d f(x) = ( i h)(ik)ei(kx ωt) dx = hkf(x) and since by the de Broglie relation hk is the momentum p of the particle, we have ˆpf(x) = pf(x) Note that this explains the choice of sign in the definition of the momentum operator! 2

3 1.3 Linear operators. An operator Â is said to be linear ifâ(cf(x)) = câf(x) and Â(f(x) + g(x)) = Âf(x) + Âg(x) where f(x) and g(x) are any two appropriate functions and c is a complex constant. Examples: the operators ˆx, ˆp and Ĥ are all linear operators. This can be checked by explicit calculation (Exercise!). 1.4 Hermitian operators. The operator Â is called the hermitian conjugate of Â if (Â ψ ) ψdx = ψ Âψdx Note: another name for hermitian conjugate is adjoint. The operator Â is called hermitian if (Âψ ) ψ dx = Examples: (i) the operator ˆx is hermitian. Indeed: (ˆxψ) ψ dx = (xψ) ψ dx = ψ Âψ dx ψ xψ dx = ψ ˆxψ dx (ii) the operator ˆp = i hd/dx is hermitian: ( (ˆpψ) ψ dx = i h dψ ) ψ dx dx ( ) dψ = i h ψ dx dx and after integration by parts, and recognizing that the wfn tends to zero as x, we get on the right-hand side i h ψ dψ dx dx = ψ ˆpψ dx (iii) the K.E. operator ˆT = ˆp 2 /2m is hermitian: ( ) 1 ˆT ψ ψ dx = (ˆp 2 ψ ) ψ dx 2m = 1 (ˆpψ) ˆpψ dx 2m = 1 ψ ˆp 2 ψ dx 2m = ψ ˆT ψ dx 3

4 (iv) the Hamiltonian is hermitian: Ĥ = ˆT + ˆV (ˆx) here ˆV is a hermitian operator by virtue of being a function of the hermitian operator ˆx, and since ˆT has been shown to be hermitian, so Ĥ is also hermitian. Theorem: The eigenvalues of hermitian operators are real. Proof: Let ψ be an eigenfunction of Â with eigenvalue a: Âψ = aψ then we have (Âψ ) ψ dx = (aψ) ψ dx = a ψ ψ dx and by hermiticity of Â we also have (Âψ ) ψ dx = ψ Âψ dx = a ψ ψ dx hence (a a) ψ ψ dx = 0 and since ψ ψ dx 0, we get a a = 0 The converse theorem also holds: an operator is hermitian if its eigenvalues are real. The proof is left as an exercise. Note: by virtue of the above theorems one can define a hermitian operator as an operator with all real eigenvalues. Corollary: The eigenvalues of the Hamiltonian are real. In fact, since by definition the Hamiltonian has the dimension of energy, therefore the eigenvalues of the Hamiltonian are the energies of the system described by the wave function. 1.5 Expectation values. Consider a system of particles with wave function ψ(x) (x can be understood to stand for all degrees of freedom of the system; so, if we have a system of two particles then x should represent {x 1, y 1, z 1 ; x 2, y 2, z 2 }). The expectation value of an operator Â that operates on ψ is defined by Â ψ Âψ dx If ψ is an eigenfunction of Â with eigenvalue a, then, assuming the wave function to be normalized, we have Â = a 4

5 Now consider the rate of change of the expectation value of Â: d Â dt = = = ( ψ Â ψ ) dx t { ψ Â t t Âψ + ψ Â t ψ + ψ Â ψ dx t {(Ĥψ ) Âψ ψ ÂĤψ} dx + ī h } where we have used the Schrödinger equation Now by hermiticity of Ĥ we get on the r.h.s.: i h ψ t = Ĥψ i {ψ h ĤÂψ ψ ÂĤψ} dx = ī ψ ( ĤÂ h ÂĤ) ψ dx hence d Â dt = Â + ī [Ĥ, Â] t h Of particular interest in applications are linear hermitian operators that do not explicitly depend on time, i.e. such that Â/ t = 0 For this class of operators we get the following equation of motion: i h d Â dt = [Â, Ĥ] Here the expectation values are taken with arbitrary square integrable functions. Therefore we can re-write this equation as an operator equation: i h dâ dt = [Â, Ĥ] If in particular Â is an observable that commutes with Ĥ, i.e. if [Â, Ĥ] = 0, then dâ dt = 0 i.e. Â is a conserved observable. We can also prove the following theorem: if two operators Â and ˆB commute, then they have common eigenfunctions. Proof: Let ψ be an eigenfunction of Â with eigenvalue a: Âψ = aψ 5

6 operating on both sides with ˆB we get ˆB(Âψ) = a ˆBψ on the l.h.s. we can write ˆBÂψ, and then since by assumption Â and ˆB commute, we get Â ˆBψ = a ˆBψ thus ˆBψ is an eigenfunction of Â with the same eigenvalue as ψ; therefore ˆBψ can differ from ψ only by a constant factor, i.e. we must have i.e. ψ is also an eigenfunction of ˆB. ˆBψ = bψ The converse theorem is also true but not as useful; I shall therefore omit the proof. 1.6 Angular momentum. Often operators can be constructed by taking the corresponding dynamical variable of classical mechanics, which is expressed in terms of coordinates and momenta, and replacing x by ˆx, p by ˆp etc. That was in fact the way we have constructed the Hamiltonian. Now we apply this prescription to angular momentum. In classical mechanics one defines the angular momentum by L = r p We get the angular momentum operator by replacing the vector r by the vector operator ˆr = (ˆx, ŷ, ẑ) and the momentum vector by the momentum vector operator where x = / x etc. ˆp = i h = i h( x, y, z ) The complete fundamental commutation relations of the coordinate and momentum operators are [ˆx, ˆp x ] = [ŷ, ˆp y ] = [ẑ, ˆp z ] = i h and [ˆx, ˆp y ] = [ˆx, ˆp z ] =... = [ẑ, ˆp y ] = 0 It will be convenient to use the following notation: ˆx 1 = ˆx, ˆx 2 = ŷ, ˆx 3 = ẑ and ˆp 1 = ˆp x, ˆp 2 = ˆp y, ˆp 3 = ˆp z we can then summarize the fundamental commutation relations by where δ ij is the Kronecker symbol: δ ij = [ˆx i, ˆp j ] = i hδ ij { 1 if i = j 0 if i j 6

7 We can now find the commutation relations for the components of the angular momentum operator. To do this it is convenient to get at first the commutation relations with ˆx i, then with ˆp i, and finally the commutation relations for the components of the angular momentum operator. Thus consider the commutator [ˆx, ˆL x ]: we have ˆL x = ŷˆp z ẑ ˆp y, and hence by the fundamental commutation relations [ˆx, ˆL x ] = 0 Next consider [ˆx, ˆL y ]: we have hence and similarly ˆL y = ẑ ˆp x ˆxˆp z [ˆx, ˆL y ] = [ˆx, ẑ ˆp x ] [ˆx, ˆxˆp z ] = i hẑ [ˆx, ˆL z ] = i hŷ etc. We can summarize the nine commutation relations: [ˆx i, ˆL j ] = i hε ijkˆx k where 1 if (ijk) = (1, 2, 3) or (2, 3, 1) or (3, 1, 2) ε ijk = 1 if (ijk) = (1, 3, 2) or (3, 2, 1) or (2, 1, 3) 0 if i = j or i = k or j = k and summation over the repeated index k is implied. Similarly one can show [ˆp i, ˆL j ] = i hε ijk ˆp k after which it is straight forward to deduce: [ˆL i, ˆL j ] = i hε ijk ˆLk The important conclusion from this result is that the components of angular momentum have no common eigenfunctions. Of course, we must also show that the angular momentum operators are hermitian. This is of course plausible (reasonable) since we know that the angular momentum is a dynamical variable in classical mechanics. The proof is left as an exercise. We can construct one more operator that commutes with all components of ˆL: define the square of ˆL by ˆL 2 = ˆL 2 x + ˆL 2 y + ˆL 2 z then [ˆL x, ˆL 2 ] = [ˆL x, ˆL 2 x + ˆL 2 y + ˆL 2 z ] = [ˆL x, ˆL 2 y] + [ˆL x, ˆL 2 z] 7

8 Now there is a simple technique to evaluate a commutator like [ ˆL x, ˆL 2 y]: write down explicitly the known commutator [ ˆL x, ˆL y ]: ˆL x ˆLy ˆL y ˆLx = i hˆl z multiply this on the left by ˆL y, then multiply on the right by ˆL y : ˆL y ˆLx ˆLy ˆL 2 y ˆL x = i hˆl y ˆLz ˆL x ˆL2 y ˆL y ˆLx ˆLy = i hˆl z ˆLy and if we add these commutation relations we get ˆL x ˆL2 y ˆL 2 y ˆL x = i h(ˆl y ˆLz + ˆL z ˆLy ) and similarly hence and similarly ˆL x ˆL2 z ˆL 2 z ˆL x = i h(ˆl y ˆLz + ˆL z ˆLy ) [ˆL x, ˆL 2 ] = 0 [ˆL y, ˆL 2 ] = [ˆL z, ˆL 2 ] = 0 Finally one can also show that the components of ˆL and ˆL 2 commute with ˆp 2, and therefore also with the K.E. operator ˆT, and that they commute with r and hence with any function of r. The latter statement is most easily shown by working in spherical polar coordinates (r, θ, ϕ), where θ is the polar angle and ϕ the azimuth. If we choose the polar axis along the cartesian z direction, then we get after some tedious calculation the following expressions for the angular momentum components: ˆL x = i h ˆL y = i h ( ( ˆL z = i h ϕ sin ϕ θ cos ϕ θ ) + cot θ cos ϕ ϕ + cot θ sin ϕ ϕ ) We can therefore conclude that the angular momentum operators commute with the Hamiltonian of a particle in a central field, for example a Coulomb field, and that implies that ˆL 2 and one of the components can be chosen to have common eigenfunctions with the Hamiltonian. 8

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