This set of Physical Chemistry Multiple Choice Questions & Answers (MCQs) focuses on “Quantum Theory Operators”.
1. The operator A> acts linearly on the functions f1 and f2. Applying it on f1 + f2 does which of the following?
a) A> (f1 + f2) = A> f1 + A> f2
b) A> (f1 + f2) = A> f1 – A> f2
c) A> (f1 + f2) = A> f1 + A> f2 + A> f1f2
d) A> (f1 + f2) = A> f1 + A> f2 – A> f1f2
View Answer
Explanation: Since A> is a linear operator, it acts on each individual function linearly, without the creation of additional terms. A> would operate separately on f1 and f2 as per this condition.
2. What is the result of applying operator A> = \(\frac{d^2}{dx^2}\) to the function f(x) = 10x3?
a) 30x2
b) 30x3
c) 60x2
d) 60x
View Answer
Explanation: Differentiating the given function once gives 30x2. Differentiating this again gives 60x. The power rule of differentiation is used here, and the given operator is a differential operator.
3. The Hamiltonian operator is the sum of all potential and kinetic energy operators in a wavefunction. How is this Hamiltonian represented in a 3-dimensional space?
a) H = V(x,y,z)
b) H = V(x,y,z) + \(\frac{-\hbar^2}{2m}\) ∇2
c) H = \(\frac{-\hbar^2}{2m}\) ∇2
d) H = V(x) + \(\frac{-\hbar^2}{2m}\) ∇2
View Answer
Explanation: In terms of operators, potential energy is written the same way as in classical mechanics. For kinetic energy, this is the momentum operator (p) divided by 2m times the Laplacian operator (for 3-D space). This is the Hamiltonian operator in 3-D space.
4. Is the function f = 8e5x an eigenfunction of the operator \(\frac{d}{dx}\). If so, what is its eigenvalue?
a) It is not an eigenfunction of the given operator
b) It is an eigenfunction with the eigenvalue 8
c) It is an eigenfunction with the eigenvalue 25
d) It is an eigenfunction with the eigenvalue 5
View Answer
Explanation: Differentiating the given function by quotient rule removes multiplies the function by 5. Dividing this new function by the original function gives an eigenvalue of 5.\(\frac{df}{dx}\) ÷ f = \(\frac{5×8e^{5x}}{8e^{5x}}\) = 5.
5. The linear momentum operator in 1-D space is given by p = –iħ \(\frac{\partial}{\partial x}\) and the kinetic energy operator is given by T = \(\frac{p^2}{2m}\). What is a simplified form of the kinetic energy operator?
a) \(\frac{\hbar}{2mx}\)
b) –\(\frac{\hbar^2}{2m} \frac{\partial}{\partial x^2}\)
c) –\(\frac{\partial}{\partial x^2}\)
d) -ħ2
View Answer
Explanation: Replacing p in the kinetic operator gives T = \(\frac{(-i\hbar \frac{\partial}{\partial x}^2)}{2m}\) = –\(\frac{\hbar^2}{2m} \frac{\partial}{\partial x^2}\). If this is extended into three dimensions, there will be additional derivatives in the y and z direction. This can be replaced by the Laplacian operator. A more generalized form of this operator is T = –\(\frac{\hbar^2}{2m}\) ∇2.
6. Do the operators for position (x) and momentum (p) commute (for given wavefunction φ)? If not, what is the operator representing the commutator between these two operators?
a) They do not commute; commuting operator is imω
b) The operators commute
c) They do not commute; commuting operator is iħω
d) They do not commute; commuting operator is iħ
View Answer
Explanation: Math shown below for this commutation operation:
[x,p]φ = x(pφ) – p(xφ) = -iħx \(\frac{d\varphi}{dx}\) + iħφ + iħx \(\frac{d\varphi}{dx}\) = iħφ.
7. What is the quantum mechanical postulate regarding the average value <a> of an observable corresponding with an operator A>?
a) < a > = ∫ φ* A> φdτ
b) < a > = ∫ (φ)* A> φ2dτ
c) < a > = ∫ φ* φdτ
d) < a > = ∫ A> dτ
View Answer
Explanation: This postulate of quantum mechanics describes the average of an observable corresponding to an operator as the integral over all space of the operator acting on the wavefunction multiplied by the complex conjugate wavefunction.
8. Is the wavefunction φ(x) = cos kx an eigenfunction of the linear momentum operator p? If yes, what is the eigenvalue?
a) No, it is not an eigenfunction
b) Yes, it is an eigenfunction with eigenvalue iħ
c) Yes, it is an eigenfunction with eigenvalue iħmω
d) Yes, it is an eigenfunction with eigenvalue iħm
View Answer
Explanation: pφ(x) = -iħ\(\frac{∂(cos(kx)}{∂x}\) = iħk sin kx : This is not a constant and hence the given wavefunction is not an eigenfunction of the linear momentum operator. It thus cannot have any eigenvalue.
9. Is the wavefunction φ(x) = eikx an eigenfunction of the kinetic energy operator T? If yes, what is its eigenvalue?
a) No, it is not an eigenfunction
b) Yes, it is an eigenfunction with eigenvalue \(\frac{\hbar^2 k^2}{2m}\)
c) Yes, it is an eigenfunction with eigenvalue \(\frac{\hbar^2}{2m}\)
d) Yes, it is an eigenfunction with eigenvalue iħm
View Answer
Explanation: Tφ(x) = –\(\frac{\hbar^2}{2m} \frac{\partial^2}{\partial x^2}\)(eikx) = \(\frac{\hbar^2 k^2}{2m}\) φ(x) => φ(x) is an eigenfunction of the kinetic energy operator with an eigenvalue \(\frac{\hbar^2 k^2}{2m}\).
10. What is the meaning of a Hermitian operator?
a) An operator that yields complex eigenvalues
b) Another name for the Hamiltonian operator
c) An operator that yields real eigenvalues
d) An operator that gives an average value over space and time
View Answer
Explanation: Hermitian operators give eigenvalues that are observable and measured quantities. These must be real to be measurable. Hermitian operators (A>) have the following property (for two defined wavefunctions Φ & φ) : ∫ φ* A> φ dτ = ∫ φ(A>φ)*dτ.
11. What is the classical expression for angular momentum (L)?
a) L = I × ω
b) L = r × F
c) L = F × p
d) L = r × p
View Answer
Explanation: Angular momentum is the rotational equivalent of linear momentum. It is defined as the cross product between a body’s linear momentum and position vector, i.e. L = r × p. L = Iω is also another way to express this quantity, but writing it as a cross product is incorrect.
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