2025-05-30
Solutions to Principles of Quantum Mechanics
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Chapter 10 Systems with NN Degrees of Freedom

10.1 NN Particles in One Dimension

Exercise 10.1.1 Show the following:

(1)

[Ω1(1)I(2),I(1)Λ2(2)]=0 for any Ω1(1) and Λ2(2)(10.1.17a)[\Omega_{1}^{(1)}\otimes I^{(2)}, I^{(1)}\otimes \Lambda_{2}^{(2)}]=0~\text{for any}~\Omega_{1}^{(1)}~\text{and}~\Lambda_{2}^{(2)}\tag{10.1.17a}

(2)

(Ω1(1)Γ2(2))(θ1(1)Λ2(2))=(Ωθ)1(1)(ΓΛ)2(2)(10.1.17b)(\Omega_{1}^{(1)}\otimes\Gamma_{2}^{(2)})(\theta_{1}^{(1)}\otimes\Lambda_{2}^{(2)})=(\Omega\theta)_{1}^{(1)}\otimes(\Gamma\Lambda)_{2}^{(2)}\tag{10.1.17b}

(3)If

[Ω1(1),Λ1(1)]=Γ1(1)[\Omega_{1}^{(1)},\Lambda_{1}^{(1)}]=\Gamma_{1}^{(1)}

then

[Ω1(1)(2),Λ1(1)(2)]=Γ1(1)I(2)(10.1.17c)[\Omega_{1}^{(1)\otimes(2)},\Lambda_{1}^{(1)\otimes(2)}]=\Gamma_{1}^{(1)}\otimes I^{(2)}\tag{10.1.17c}

and similarly with 121\to 2.

(4)

(Ω1(1)(2)+Ω2(1)(2))2=(Ω12)(1)I(2)+I(1)(Ω22)(2)+2Ω1(1)Ω2(2)(10.1.17d)(\Omega_{1}^{(1)\otimes (2)}+\Omega_{2}^{(1)\otimes(2)})^{2}=(\Omega_{1}^{2})^{(1)}\otimes I^{(2)}+I^{(1)}\otimes (\Omega_{2}^{2})^{(2)}+2\Omega_{1}^{(1)}\otimes\Omega_{2}^{(2)}\tag{10.1.17d}
2025-05-30
Solutions to Principles of Quantum Mechanics
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Chapter 9 The Heisenberg Uncertainty Relations

9.1 Introduction

9.2 Derivation of the Uncertainty Relations

9.3 The Minimum Uncertainty Packet

9.4 Applications of the Uncertainty Principle

Exercise 9.4.1 Consider the oscillator in the state n=1|n=1\rangle and verify that

1X21X2mω\left\langle\frac{1}{X^2}\right\rangle \simeq \frac{1}{\left\langle X^2\right\rangle} \simeq \frac{m \omega}{\hbar}
2025-05-30
Solutions to Principles of Quantum Mechanics
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Chapter 8 The Path Integral Formulation of Quantum Theory

8.1 The Path Integral Recipe

8.2 Analysis of the Recipe

8.3 An Approximation to U(t)U(t) for the Free Particle

8.4 Path Integral Evaluation of the Free-Particle Propagator

8.5 Equivalence to the Schrödinger Equation

8.6 Potentials of the Form V=a+bx+cx2+dx˙+exx˙V=a+bx+cx^{2}+d\dot{x}+ex\dot{x}

Exercise 8.6.1 Verify that

U(x,t;x,0)=A(t)exp(iScl/), A(t)=(m2πit)1/2U(x,t;x^{\prime},0)=A(t)\exp(\mathrm{i}S_{\text{cl}}/\hbar),~A(t)=\left(\dfrac{m}{2\pi\hbar\mathrm{i}t}\right)^{1/2}

agrees with the exact result, Eq. (5.4.31), for V(x)=fxV(x)=-fx. Hint: Start with xcl(t)=x0+v0t+12(f/m)t2x_{\text{cl}}(t^{\prime\prime})=x_{0}+v_{0}t^{\prime\prime}+\dfrac{1}{2}(f/m)t^{\prime\prime 2} and find the constants x0x_{0} and v0v_{0} from the requirement that xcl(0)=xx_{\text{cl}}(0)=x^{\prime} and xcl(t)=xx_{\text{cl}}(t)=x.

2025-05-30
Solutions to Principles of Quantum Mechanics
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Chapter 7 The Harmonic Oscillator

7.1 Why Study the Harmonic Oscillator?

7.2 Review of the Classical Oscillator

7.3 Quantization of the Oscillator (Coordinate Basis)

Exercise 7.3.1 Consider the question why we tried a power-series solution for Eq. (7.3.11) but not Eq. (7.3.8). By feeding in a series into the latter, verify that a three-term recursion relation between Cn+2C_{n+2}, CnC_{n}, and Cn2C_{n-2} obtains, from which the solution does not follow so readily. The problem is that ψ\psi^{\prime\prime} has two powers of yy less than 2εψ2 \varepsilon \psi, while the y2-y^{2} piece has two more powers of yy. In Eq. (7.3.11) on the other hand, of the three pieces uu^{\prime\prime}, 2yu-2yu^{\prime}, and (2ε1)u(2\varepsilon-1)u, the last two have the same powers of yy.

2025-02-19
Solutions to Principles of Quantum Mechanics
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