This set of Physical Chemistry Multiple Choice Questions & Answers (MCQs) focuses on “Quantum Theory Operators – Set 2”.

1. What is the quantum mechanical operator for angular momentum in the x-direction (L_{x})?

a) *L _{x} = -iħ (y \(\frac{\partial}{\partial z}\) – z \(\frac{\partial}{\partial y}\))*

b)

*L*

_{x}= -iħ (z \(\frac{\partial}{\partial x}\) – x \(\frac{\partial}{\partial z}\))c)

*L*

_{x}= -iħ (x \(\frac{\partial}{\partial y}\) – y \(\frac{\partial}{\partial x}\))d)

*L*

_{x}= -iħ (y \(\frac{\partial}{\partial z}\) – y \(\frac{\partial}{\partial y}\))View Answer

Explanation: Quantum mechanical angular momentum operator in a given direction is defined by the change in momentum in the other direction in a coordinate system. Hence

*L*is angular momentum in the y-direction and L

_{x}= yp_{z}– zp_{y}= -iħ (y \(\frac{\partial}{\partial z}\) – z \(\frac{\partial}{\partial y}\)). L_{x}= -iħ (z \(\frac{\partial}{\partial x}\) – x \(\frac{\partial}{\partial z}\))*is the angular momentum in the z-direction.*

_{x}= -iħ (x \(\frac{\partial}{\partial y}\) – y \(\frac{\partial}{\partial x}\))2. Which of the following operators/ quantities is not Hermitian?

a) Momentum

b) Position

c) Atomic acceleration

d) Kinetic Energy

View Answer

Explanation: A Hermitian operator is a physical observable by default. Atomic Acceleration is not a physical observable as it cannot be witnessed and measured by ordinary spectroscopic instrument. Mathematically, a Hermitian operator satisfies the following condition:

*∫ φ*.

^{*}A^{>}φ dτ = ∫ φ (A^{>}φ)^{*}dτ3. Can the overall angular momentum operator (L) for a quantum mechanical system be known accurately?

a) Yes, by taking the vector addition of L_{x}, L_{y}, L_{z}

b) No, the projection of angular momentum can be known accurately only along one axis

c) No, the projection of angular momentum can be known accurately only along two axes

d) Yes, by taking a cross product of projections of angular momentum along each axes

View Answer

Explanation: Angular momentum can only be known accurately along one axis for a given system. As a default, the L

_{z}axis is chosen and defined. Quantum mechanics does not allow more than one projection, and hence overall angular momentum to be accurately defined.

4. In general, what property does the multiplication of operators obey?

a) Associative

b) Commutative

c) Parallelism

d) Orthogonality

View Answer

Explanation: The multiplication of operators is always associative, i.e. for three operators A, B, and C, ABC = (AB)C = A(BC). Operators may follow the commutative property, but this is not always the case.

5. What is the meaning of the commutator of two operations A and B?

a) [A,B] = AA-BB

b) [A,B] = BA-AB

c) [A,B] = AB-BA

d) [A,B] = BB-AA

View Answer

Explanation: The commutator is defined as [A,B] = AB-BA. In this case, the order of applying the operations to the given function matters and changing the order can give drastically different results. The commutator can tell us various things about the operations being applied and their implications on different wavefunctions.

6. In what special case does the order of operations not matter?

a) [A,B] = ∞

b) [A,B] = -∞

c) [A,B] = 0

d) [A,B] is not defined

View Answer

Explanation: When [A,B] = 0, the order of operations does not matter as AB = BA in that case. This directly equates two different orders of operation. If [A,B] = ±∞ or is not defined, then the order of operations severely affects the two operations.

7. Operation A = x and B = d/dx are applied on an arbitrary function f. Are these two operators commutative?

a) Yes, they commute

b) No, they do not commute

c) Their commutator depends on the nature of the function

d) The algebra with operator B only works on well defined functions

View Answer

Explanation:

*[A,B] = ABf – BAf = xf′ – xf′ – f ≠*0. Hence the two operators do not commute. The commutator does not generally depend on the nature of the function. If the algebra works out without a well-defined function, then any operator can work.

8. Operation A = x and B = d/dx are applied on an arbitrary function f. What is the commutator for these two operations on f?

a) 0

b) -1

c) 1

d) ∞

View Answer

Explanation:

*[A,B] = ABf – BAf = xf′ – xf′ – f = -f*. Hence the commutator is -1 as the function changes sign upon applying the commutator function.

9. What is the quantum mechanical operator for position x?

a) x

b) \(\frac{∂}{∂x}\)

c) iħ \(\frac{∂}{∂x}\)

d) -iħ \(\frac{∂}{∂x}\)

View Answer

Explanation: The quantum mechanical operator and classical definition of position x performs the same operation in classical as well as quantum mechanics. The given function is simply multiplied by x. This is no inclusion of Plank’s constant, imaginary numbers, or partial derivatives.

10. The below graph is a sketch of the wavefunction ψ(x) = e^{-3x2} sin(2x). Is the function even or odd, and what is the subsequent overall integral over all space?

a) function is even, integral over all space = 0

b) function is odd, integral over all space = 0

c) function is even, integral over all space = 2π^{2}

d) function is odd, integral over all space = 2π^{2}

View Answer

Explanation: The criteria for an odd function is f(-x) = -f(x). For this wavefunction, f(-x) = e

^{-3(-x)2}sin(2(-x)) = e

^{-3x2}sin(2x), since sin(x) is odd. Similarly, -f(x) = -e

^{-3x2}sin(2x). Since this satisfies the criteria for an odd function, the wavefunction is odd. The integral over all space for an odd function is always zero, since it is symmetric about the origin, hence the positive and negative parts cancel out.

11. Is the wavefunction ψ(x) = e^{-3x2} sin(2x) an eigenfunction of the operator A = d/dx? If yes, what is its eigenvalue?

a) No, the given function is not an eigenfunction of A

b) Yes, the given function is an eigenfunction, with eigenvalue = -1

c) Yes, the given function is an eigenfunction, with eigenvalue = 1

d) Yes, the given function is an eigenfunction, with eigenvalue = 2

View Answer

Explanation: \(\frac{d}{dx}\)(e

^{-3x2}sin(2x)) = 2e

^{-3x2}cos(2x) – 6xe

^{-3x2}sin(2x)—→ hence not an eigenfunction as differentiation just creates additional terms as per the chain and product rule of differentiation.

**Sanfoundry Global Education & Learning Series – Physical Chemistry.**

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