Question

qm 09.3

3. An operator  is Hermitian if it satisfies the condition $ $(y) dx = (Ap) u dx, for any wavefunctions $(x) and y(x). (i) The time dependent Schrödinger equation is ih au = fu, at where the Hamiltonian operator is Hermitian. Show that the equation of mo- tion for the expectation value of any Hermitian operator  is given by d(A) IH, Â]), dt ħi i = where the operator  does not depend explicitly on time. [7 marks) (ii) The Hamiltonian operator is the sum of the kinetic energy operator Î and the potential energy operator Û = V(r). In three dimensions, the x component orbital angular momentum operator is Lx = ģP2 – žby. (a) Using the explicit form for the components of the momentum operator P = -ih, [7 marks] show that Îx commutes with Î. (b) Show that the commutator of £x with û is given by av [û îx] = iħ y az Z av ay [3 marks] (c) Without further calculation, generalise the expression for [Û, 1x] to a vector equation for [W, Î). Comment on the resulting equation for the time depen- dence of the expectation value of the orbital angular momentum with regard to the equivalent classical expression. [3 marks] [Total 20 marks]

Answer 1

3) i The Hermitian operator can be expressed as: $\langle\psi|A\psi\rangle=\langle A \psi|\psi\rangle$

So,

მu' (A) dt d (VA) dt au at |40) + (v ӘA at +(142 at

and from Schrodinger equation: $\frac{\partial \psi}{\partial t}=\frac{1}{i\hbar}H\psi$ . and $\frac{\partial A}{\partial t}=0$ since it is explicitly independent.

So, $\frac{d\langle A\rangle}{dt}=-\frac{1}{i\hbar}\langle H\psi|A\psi\rangle+\frac{1}{i\hbar}\langle \psi |AH\psi\rangle$

But as is hermitain we can use the above equation:

$\frac{d\langle A\rangle}{dt}=\frac{i}{\hbar}[\langle \psi|HA\psi\rangle-\langle \psi |AH\psi\rangle]$ Since is also hermitian.

$\frac{d\langle A\rangle}{dt}=\frac{i}{\hbar}\langle [H,A]\rangle$

ii) a) $[L_x,p]=-\frac{i\hbar}{2m}(i\hbar)^2\Big[y\frac{\partial}{\partial z}-z\frac{\partial}{\partial y},\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}+\frac{\partial^2}{\partial z^2}\Big]=0$

Since $\Big[\frac{\partial^2}{\partial x^2},\frac{\partial}{\partial z}\Big]=\Big[\frac{\partial^2}{\partial y^2},\frac{\partial^2}{\partial z}\Big]=\Big[\frac{\partial}{\partial z^2},\frac{\partial}{\partial z}\Big]=0$ and similarly for y.

b) $[V,L_x]=-i\hbar \Big[V,y\frac{\partial}{\partial z}-y\frac{\partial}{\partial z}\Big]=i\hbar \Big[y\frac{\partial}{\partial z}-y\frac{\partial}{\partial z},V\Big]=i\hbar \Big[y\frac{\partial V}{\partial z}-y\frac{\partial V}{\partial z}\Big]$

Since, $[A,B]=-[B,A]$