Operators and Observables Questions and Answers

Q: A particle is in the normalized state \(|\psi\rangle\). \(|\psi\rangle\) is a superposition of energy eigen states \(|E_{0} = 10\ \text{eV}\rangle\) and \(|E_{1} = 30\ \text{eV}\rangle\). The average value of energy of the particle in the state \(|\psi\rangle\) is 20 eV. The state \(|\psi\rangle\) is :

The state is \(\frac{1}{\sqrt{2}}|E_{0}\rangle - \frac{1}{\sqrt{2}}|E_{1}\rangle\) (option D).Write \(|\psi\rangle = c_{0}|E_{0}\rangle + c_{1}|E_{1}\rangle\). Normalisation and the given average energy require:\[ \begin{aligned} |c_{0}|^{2} + |c_{1}|^{2} &= 1 \\ 10|c_{0}|^{2} + 30|c_{1}|^{2} &= 20 \end{aligned} \]so \(|c_{0}|^{2} = |c_{1}|^{2} = 1/2\), and \(|c_{0}| = |c_{1}| = 1/\sqrt{2}\). The relative sign is a phase and does not affect the average energy, so option D fits. Options A and C are not normalised (\(1/4 + 3/16 = 7/16\)), and option B is normalised but gives \(10/3 + 20 = 23.3\) eV.

Q: For any operator A, i (A+ - A) is :

The operator \(i(A^{+} - A)\) is Hermitian.Take the Hermitian conjugate, using \((cX)^{+} = c^{*}X^{+}\) and \((A^{+})^{+} = A\):\[ \begin{aligned} [i(A^{+} - A)]^{+} &= -i(A - A^{+}) \\ &= i(A^{+} - A) \end{aligned} \]The operator equals its own conjugate, which is the definition of Hermitian. The bracket \(A^{+} - A\) alone is anti-Hermitian, and multiplying an anti-Hermitian operator by i makes it Hermitian. Unitarity (\(U^{+}U = 1\)) and orthogonality are different properties that do not follow for an arbitrary A.

Q: If quantum mechanical operators of two observables of a system do not commute, then :

If the operators do not commute, it is impossible to know the precise values of the observables simultaneously.Two observables can be measured simultaneously with arbitrary precision only if they share a common set of eigenstates, which is the case exactly when their operators commute. For non-commuting operators the generalised uncertainty principle sets a lower limit:\[ \begin{aligned} \Delta A\, \Delta B &\ge \frac{1}{2}\left|\langle C \rangle\right| \\ C &= [\hat{A}, \hat{B}] \end{aligned} \]Position and momentum, with \([x, p] = i\hbar\), are the standard example. Commutation says nothing about the sign of the energy, the parity of the wave function, or whether the observables depend on time, so the other options are unrelated.

Q: If \(\bar{A}\) and \(\bar{B}\) are quantum mechanical operators, then \([[\bar{A}, \bar{B}], [\bar{B}, \bar{A}]]\) is equal to :

The commutator is 0.Let \(C = [\bar{A}, \bar{B}]\). Then \([\bar{B}, \bar{A}] = -C\), and\[ [C, -C] = -(CC - CC) = 0 \]since every operator commutes with itself and with any multiple of itself. This holds for any pair of operators.The first option is what you get by expanding carelessly: it is not identically zero in general, so it cannot equal a quantity that always vanishes. The third option and \([A, B]^{2}\) are likewise non-zero for general operators, for example position and momentum, where \([x, p]^{2} = -\hbar^{2}\).

Q: Eigenvalues of the sum of two operators is the sum of the eigenvalues if :

The eigenvalues add when the operators have diagonalizable commuting matrices.If \(A\) and \(B\) are diagonalizable and commute, they can be diagonalized by the same basis. In that basis each common eigenvector \(v\) satisfies \(Av = av\) and \(Bv = bv\), so \((A + B)v = (a + b)v\): the eigenvalues of the sum are sums of corresponding eigenvalues.Being Hermitian, unitary or anti-Hermitian is not enough. For example, the Pauli matrices \(\sigma_{x}\) and \(\sigma_{z}\) are Hermitian with eigenvalues \(\pm 1\), but \(\sigma_{x} + \sigma_{z}\) has eigenvalues \(\pm\sqrt{2}\), because they do not commute.

Q: If H is a Hermitian operator, the following is "always" true :

It is always true that H has real diagonal matrix elements.A Hermitian operator satisfies \(H_{ij} = H_{ji}^{*}\). Setting \(i = j\) gives \(H_{ii} = H_{ii}^{*}\), so every diagonal element is real, in any basis.The others are not guaranteed. Diagonal elements can be negative (for example \(-\sigma_{z}\)). Off-diagonal elements are complex in general, and can be real, as in \(\sigma_{x}\). The trace is the sum of real eigenvalues and need not be zero (the identity is Hermitian).

Q: The operator \(\left(\frac{d}{dx} - x\right)\left(\frac{d}{dx} + x\right)\) is equivalent to :

The operator is \(\frac{d^{2}}{dx^{2}} - x^{2} + 1\).Act on a test function \(f\), remembering that \(d/dx\) also differentiates the \(x\) in \(xf\):\[ \begin{aligned} \left(\frac{d}{dx} - x\right)(f' + xf) &= f'' + (xf)' - xf' - x^{2}f \\ &= f'' + f + xf' - xf' - x^{2}f \\ &= f'' - x^{2}f + f \end{aligned} \]The extra 1 comes from the commutator \([d/dx, x] = 1\). This is the same algebra that gives the \(\pm\frac{1}{2}\hbar\omega\) in the harmonic-oscillator ladder operators.\(d^{2}/dx^{2} - x^{2}\) treats the two factors as commuting numbers and drops the commutator. The options containing a first-derivative term keep the \(xf'\) pieces, which actually cancel.

11.

A particle is in the normalized state \(|\psi\rangle\). \(|\psi\rangle\) is a superposition of energy eigen states \(|E_{0} = 10\ \text{eV}\rangle\) and \(|E_{1} = 30\ \text{eV}\rangle\). The average value of energy of the particle in the state \(|\psi\rangle\) is 20 eV. The state \(|\psi\rangle\) is :

A.\(\frac{1}{2}|E_{0} = 10\,\text{eV}\rangle\) + \(\frac{\sqrt{3}}{4}|E_{1} = 30\,\text{eV}\rangle\)
B.\(\frac{1}{\sqrt{3}}|E_{0} = 10\,\text{eV}\rangle\) + \(\sqrt{\frac{2}{3}}|E_{1} = 30\,\text{eV}\rangle\)
C.\(\frac{1}{2}|E_{0} = 10\,\text{eV}\rangle\) - \(\frac{\sqrt{3}}{4}|E_{1} = 30\,\text{eV}\rangle\)
D.\(\frac{1}{\sqrt{2}}|E_{0} = 10\,\text{eV}\rangle\) - \(\frac{1}{\sqrt{2}}|E_{1} = 30\,\text{eV}\rangle\)

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: D

The state is \(\frac{1}{\sqrt{2}}|E_{0}\rangle - \frac{1}{\sqrt{2}}|E_{1}\rangle\) (option D).

Write \(|\psi\rangle = c_{0}|E_{0}\rangle + c_{1}|E_{1}\rangle\). Normalisation and the given average energy require:

\[ \begin{aligned} |c_{0}|^{2} + |c_{1}|^{2} &= 1 \\ 10|c_{0}|^{2} + 30|c_{1}|^{2} &= 20 \end{aligned} \]

so \(|c_{0}|^{2} = |c_{1}|^{2} = 1/2\), and \(|c_{0}| = |c_{1}| = 1/\sqrt{2}\). The relative sign is a phase and does not affect the average energy, so option D fits. Options A and C are not normalised (\(1/4 + 3/16 = 7/16\)), and option B is normalised but gives \(10/3 + 20 = 23.3\) eV.

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12.

For any operator A, i (A⁺ - A) is :

A.Hermitian
B.Anti-Hermitian
C.Unitary
D.Orthogonal

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: A

The operator \(i(A^{+} - A)\) is Hermitian.

Take the Hermitian conjugate, using \((cX)^{+} = c^{*}X^{+}\) and \((A^{+})^{+} = A\):

\[ \begin{aligned} [i(A^{+} - A)]^{+} &= -i(A - A^{+}) \\ &= i(A^{+} - A) \end{aligned} \]

The operator equals its own conjugate, which is the definition of Hermitian. The bracket \(A^{+} - A\) alone is anti-Hermitian, and multiplying an anti-Hermitian operator by i makes it Hermitian. Unitarity (\(U^{+}U = 1\)) and orthogonality are different properties that do not follow for an arbitrary A.

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13.

If quantum mechanical operators of two observables of a system do not commute, then :

A.the total energy of the system must be negative
B.the parity of the wave function of the system must be odd
C.the observables corresponding to these operators must be time-independent
D.it is impossible to know the precise values of the observables simultaneously

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: D

If the operators do not commute, it is impossible to know the precise values of the observables simultaneously.

Two observables can be measured simultaneously with arbitrary precision only if they share a common set of eigenstates, which is the case exactly when their operators commute. For non-commuting operators the generalised uncertainty principle sets a lower limit:

\[ \begin{aligned} \Delta A\, \Delta B &\ge \frac{1}{2}\left|\langle C \rangle\right| \\ C &= [\hat{A}, \hat{B}] \end{aligned} \]

Position and momentum, with \([x, p] = i\hbar\), are the standard example. Commutation says nothing about the sign of the energy, the parity of the wave function, or whether the observables depend on time, so the other options are unrelated.

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14.

If \(\bar{A}\) and \(\bar{B}\) are quantum mechanical operators, then \([[\bar{A}, \bar{B}], [\bar{B}, \bar{A}]]\) is equal to :

A.\(\bar{A}\bar{B}\bar{B}\bar{A} - \bar{B}\bar{A}\bar{A}\bar{B}\)
B.0
C.\(\bar{A}\bar{A}\bar{B} - \bar{B}\bar{B}\bar{A}\)
D.\([A, B]^{2}\)

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: B

The commutator is 0.

Let \(C = [\bar{A}, \bar{B}]\). Then \([\bar{B}, \bar{A}] = -C\), and

\[ [C, -C] = -(CC - CC) = 0 \]

since every operator commutes with itself and with any multiple of itself. This holds for any pair of operators.

The first option is what you get by expanding carelessly: it is not identically zero in general, so it cannot equal a quantity that always vanishes. The third option and \([A, B]^{2}\) are likewise non-zero for general operators, for example position and momentum, where \([x, p]^{2} = -\hbar^{2}\).

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15.

Eigenvalues of the sum of two operators is the sum of the eigenvalues if :

A.The operators are Hermitian
B.The operators have diagonalizable commuting matrices
C.The operators are unitary
D.The operators are anti-Hermitian

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: B

The eigenvalues add when the operators have diagonalizable commuting matrices.

If \(A\) and \(B\) are diagonalizable and commute, they can be diagonalized by the same basis. In that basis each common eigenvector \(v\) satisfies \(Av = av\) and \(Bv = bv\), so \((A + B)v = (a + b)v\): the eigenvalues of the sum are sums of corresponding eigenvalues.

Being Hermitian, unitary or anti-Hermitian is not enough. For example, the Pauli matrices \(\sigma_{x}\) and \(\sigma_{z}\) are Hermitian with eigenvalues \(\pm 1\), but \(\sigma_{x} + \sigma_{z}\) has eigenvalues \(\pm\sqrt{2}\), because they do not commute.

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16.

If H is a Hermitian operator, the following is "always" true :

A.H has real diagonal matrix elements
B.H has positive diagonal matrix elements
C.H has imaginary off-diagonal matrix elements
D.H has a trace equal to zero

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: A

It is always true that H has real diagonal matrix elements.

A Hermitian operator satisfies \(H_{ij} = H_{ji}^{*}\). Setting \(i = j\) gives \(H_{ii} = H_{ii}^{*}\), so every diagonal element is real, in any basis.

The others are not guaranteed. Diagonal elements can be negative (for example \(-\sigma_{z}\)). Off-diagonal elements are complex in general, and can be real, as in \(\sigma_{x}\). The trace is the sum of real eigenvalues and need not be zero (the identity is Hermitian).

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17.

The operator \(\left(\frac{d}{dx} - x\right)\left(\frac{d}{dx} + x\right)\) is equivalent to :

A.\(\frac{d^{2}}{dx^{2}} - x^{2}\)
B.\(\frac{d^{2}}{dx^{2}} - x^{2} + 1\)
C.\(\frac{d^{2}}{dx^{2}} - x\frac{d}{dx}x^{2} + 1\)
D.\(\frac{d^{2}}{dx^{2}} - 2x\frac{d}{dx} - x^{2}\)

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

Answer: B

The operator is \(\frac{d^{2}}{dx^{2}} - x^{2} + 1\).

Act on a test function \(f\), remembering that \(d/dx\) also differentiates the \(x\) in \(xf\):

\[ \begin{aligned} \left(\frac{d}{dx} - x\right)(f' + xf) &= f'' + (xf)' - xf' - x^{2}f \\ &= f'' + f + xf' - xf' - x^{2}f \\ &= f'' - x^{2}f + f \end{aligned} \]

The extra 1 comes from the commutator \([d/dx, x] = 1\). This is the same algebra that gives the \(\pm\frac{1}{2}\hbar\omega\) in the harmonic-oscillator ladder operators.

\(d^{2}/dx^{2} - x^{2}\) treats the two factors as commuting numbers and drops the commutator. The options containing a first-derivative term keep the \(xf'\) pieces, which actually cancel.

Discuss in Forum

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