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Chapter 6: Problem 7

Consider two particles of mass \(m_{1}\) and \(m_{2}\) (in one dimension) that interact via a potential that depends only on the distance between the particles \(V\left(\left|x_{1}-x_{2}\right|\right),\) so that the Hamiltonian is \(\hat{H}=-\frac{\hbar^{2}}{2 m_{1}} \frac{\partial^{2}}{\partial x_{1}^{2}}-\frac{\hbar^{2}}{2 m_{2}} \frac{\partial^{2}}{\partial x_{2}^{2}}+V\left(\left|x_{1}-x_{2}\right|\right)\) Acting on a two-particle wave function the translation operator would be \(\hat{T}(a) \psi\left(x_{1}, x_{2}\right)=\psi\left(x_{1}-a, x_{2}-a\right)\) (a) Show that the translation operator can be written \(\hat{T}(a)=e^{-\frac{i a}{\hbar} \hat{P}}\) where \(\hat{P}=\hat{p}_{1}+\hat{p}_{2}\) is the total momentum. (b) Show that the total momentum is conserved for this system.

Short Answer

Expert verified
(a) \( \hat{T}(a) = e^{-\frac{i a}{\hbar} \hat{P}} \). (b) Total momentum \( \hat{P} \) is conserved since \( [\hat{H}, \hat{P}] = 0 \).

Step by step solution

01

Express the Translation Operator

Given the translation operator \( \hat{T}(a) \), the action is defined as: \( \hat{T}(a) \psi(x_1, x_2) = \psi(x_1 - a, x_2 - a) \). This operator translates both coordinates by \(-a\).
02

Define Total Momentum Operator

The total momentum \( \hat{P} \) is defined as the sum of the individual momenta: \( \hat{P} = \hat{p}_1 + \hat{p}_2 \). Individual momentum operators are defined by \( \hat{p}_i = -i\hbar \frac{\partial}{\partial x_i} \) for \( i = 1, 2 \).
03

Exponential Form of Translation Operator

The translation operator \( \hat{T}(a) \) can be written in exponential form \( \hat{T}(a) = e^{-\frac{i a}{\hbar} \hat{P}} \). This stems from the property of exponential operators in quantum mechanics: translation in space corresponds to a shift in momentum space.
04

Verify Exponential Form

Use the fact: \( e^{-\frac{i a}{\hbar} \hat{P}} \psi(x_1, x_2) = \psi(x_1 - a, x_2 - a) \) can be derived from \( \psi(x_i - a) \approx \psi(x_i) - a \frac{\partial}{\partial x_i} \psi(x_i) = \psi(x_i) + \frac{i a}{\hbar} \hat{p}_i \psi(x_i) \) for a small \( a \).
05

Show Conservation鈥擳otal Momentum Hamiltonian Commutation

To show total momentum conservation, check if \( [\hat{H}, \hat{P}] = 0 \). Using the Hamiltonian \( \hat{H} \) and \( \hat{P} = \hat{p}_1 + \hat{p}_2 \), the commutator evaluates as: \[ [\hat{H}, \hat{P}] = \left[-\frac{\hbar^2}{2m_1}\frac{\partial^2}{\partial x_1^2} - \frac{\hbar^2}{2m_2}\frac{\partial^2}{\partial x_2^2} + V(|x_1-x_2|), \hat{p}_1 + \hat{p}_2 \right] \] Using properties of differentiation and the potential's dependency, this equals zero.
06

Conclusion

Since \( [\hat{H}, \hat{P}] = 0 \), the total momentum \( \hat{P} \) commutes with the Hamiltonian, indicating it is a conserved quantity under the evolution of this system.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Two-particle Systems
In quantum mechanics, two-particle systems refer to cases where two individual particles interact with one another. These systems are fundamental in understanding more complex interactions in physical chemistry and condensed matter. For our specific scenario, imagine two particles with masses \( m_1 \) and \( m_2 \) that are restricted to move along a single line.

The interaction between these particles depends solely on the distance between them. This is represented by the potential \( V(|x_1 - x_2|) \), ensuring that the energy due to the interaction changes solely based on how far apart the particles are. Such potentials are commonly seen in real-world phenomena like gravitational and electrostatic forces, which depend on how far apart two objects or charges are.
  • Systems described in a one-dimensional space simplify analysis and help us focus on the effects of quantum interactions.
  • Understanding interactions via potentials is crucial for applications in materials science and molecular chemistry.
Momentum Conservation
Momentum conservation is a critical principle in physics, asserting that the total momentum of a closed system remains constant over time, provided no external forces act upon it. In quantum mechanics, this principle often manifests through the commutation relationship between the Hamiltonian \( \hat{H} \) and the momentum operator \( \hat{P} \).

For our two-particle system, the total momentum is given by \( \hat{P} = \hat{p}_1 + \hat{p}_2 \), which considers the contribution from each particle's momentum. The goal is to determine if this total momentum remains conserved, i.e., does not change with time while the system evolves under the given Hamiltonian.
  • If the operators \( \hat{H} \) and \( \hat{P} \) commute, then \( [\hat{H}, \hat{P}] = 0 \), meaning the total momentum is conserved.
  • A commuting relationship suggests that the physical property in question (here, momentum) does not change over time, confirming its conservation.
Understanding this principle sheds light on fundamental symmetries in physics and helps predict system behavior over time.
Translation Operator
A translation operator in quantum mechanics describes how a quantum state shifts in space. It plays an essential role in determining how wavefunctions, which describe the quantum state of particles, change when the system undergoes spatial translation.

The translation operator \( \hat{T}(a) \) acts to shift each coordinate \( x_1 \) and \( x_2 \) by \(-a\). In mathematical terms, it modifies the wavefunction as \( \hat{T}(a) \psi(x_1, x_2) = \psi(x_1 - a, x_2 - a) \). This action can be expressed in terms of the exponential form \( \hat{T}(a) = e^{-\frac{i a}{\hbar} \hat{P}} \), where \( \hat{P} \) is the total momentum operator.
  • This exponential expression links spatial translation with the momentum, illustrating how shifts in position affect momentum.
  • This is a direct consequence of the wave-like nature of particles in quantum mechanics, characterized by their wavefunctions.
This understanding bridges the gap between classical intuitive ideas of movement and the quantum mechanical treatment of particles, showing how translations can affect a system's state and momentum.

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Most popular questions from this chapter

Consider the free particle in one dimension: \(\hat{H}=\hat{p}^{2} / 2 m .\) This Hamiltonian has both translational symmetry and inversion symmetry. (a) Show that translations and inversion don't commute. (b) Because of the translational symmetry we know that the eigenstates of \(\hat{H}\) can be chosen to be simultaneous eigenstates of momentum, namely \(f_{p}(x)\) (Equation 3.32 ). Show that the parity operator turns \(f_{p}(x)\) into \(f_{-p}(x) ;\) these two states must therefore have the same energy. (c) Alternatively, because of the inversion symmetry we know that the eigenstates of \(\hat{H}\) can be chosen to be simultaneous eigenstates of parity, namely \(\frac{1}{\sqrt{\pi \hbar}} \cos \left(\frac{p x}{\hbar}\right)\) and \(\frac{1}{\sqrt{\pi \hbar}} \sin \left(\frac{p x}{\hbar}\right).\) Show that the translation operator mixes these two states together; they therefore must be degenerate. Note: Both parity and translational invariance are required to explain the degeneracy in the free-particle spectrum. Without parity, there is no reason for \(f_{p}(x)\) and \(f_{-p}(x)\) to have the same energy (I mean no reason based on symmetries discussed thus far ...obviously you can plug them in to the timeindependent Schr枚dinger equation and show it's true).

Show that, for a Hermitian operator \(\hat{Q},\) the operator \(\hat{U}=\exp [i \hat{Q}]\) is unitary. Hint: First you need to prove that the adjoint is given by \(\hat{U}^{\dagger}=\exp [-i \hat{Q}] ;\) then prove that \(\hat{U}^{\dagger} \hat{U}=1 .\) Problem \(\underline{3.5}\) may help.

Problem 6.27 Consider a free particle of mass \(m .\) Show that the position and momentum operators in the Heisenberg picture are given by \(\hat{x}_{H}(t)=\hat{x}_{H}(0)+\frac{1}{m} \hat{p}_{H}(0)\) \(\hat{p}_{H}(t)=\hat{p}_{H}(0)\) Comment on the relationship between these equations and the classical equations of motion. Hint: you will first need to evaluate the commutator \(\left[\hat{x}, \hat{H}^{n}\right] ;\) this will allow you to evaluate the commutator \([\hat{x}, \hat{U}]\)

In this problem you will establish the correspondence between Equations 6.30 and 6.31. (a) Diagonalize the matrix \(^{16}\) \(M=\left(\begin{array}{cc}1 & -\varphi / N \\ \varphi / N & 1\end{array}\right)\) to obtain the matrix \(\mathrm{M}^{\prime}=\mathrm{SMS}^{-1}\) where \(S^{-1}\) is the unitary matrix whose columns are the (normalized) eigenvectors of M. (b) Use the binomial expansion to show that \(\lim _{N \rightarrow \infty}\left(\mathrm{M}^{\prime}\right)^{N}\) is a diagonal matrix with entries \(e^{-i \varphi \text { and } e^{i \varphi} \text { on the diagonal. }}\) (c) Transform back to the original basis to show that \(\lim _{N \rightarrow \infty} \mathrm{M}^{N}=\mathrm{S}^{-1}\left[\lim _{N \rightarrow \infty}\left(\mathrm{M}^{\prime}\right)^{N}\right] \mathrm{S}\) agrees with the matrix in Equation 6.31.

Consider a time-independent Hamiltonian for a particle moving in one dimension that has stationary states \(\psi_{n}(x)\) with energies \(E_{n}\) (a) Show that the solution to the time-dependent Schr枚dinger equation can be written \(\Psi(x, t)=\hat{U}(t) \Psi(x, 0)=\int K\left(x, x^{\prime}, t\right) \Psi\left(x^{\prime}, 0\right) d x^{\prime}\) where \(K\left(x, x^{\prime}, t\right),\) known as the propagator, is \(K\left(x, x^{\prime}, t\right)=\sum_{n} \psi_{n}^{*}\left(x^{\prime}\right) e^{-i E_{n} t / \hbar} \psi_{n}(x)\) Here \(\left|K\left(x, x^{\prime}, t\right)\right|^{2}\) is the probability for a quantum mechanical particle to travel from position \(x^{\prime}\) to position \(x\) in time \(t.\) (b) Find \(K\) for a particle of mass \(m\) in a simple harmonic oscillator potential of frequency \(\omega .\) You will need the identity $$\frac{1}{\sqrt{1-z^{2}}} \exp \left[-\frac{\xi^{2}+\eta^{2}-2 \xi \eta z}{1-z^{2}}\right]=e^{-\xi^{2}} e^{-\eta^{2}} \sum_{n=0}^{\infty} \frac{z^{n}}{2^{n} n !} H_{n}(\xi) H_{n}(\eta)$$ (c) Find \(\Psi(x, t)\) if the particle from part (a) is initially in the state \(^{40}\) \(\Psi(x, 0)=\left(\frac{2 a}{\pi}\right)^{1 / 4} e^{-a\left(x-x_{0}\right)^{2}}\) Compare your answer with Problem \(2.49 .\) Note: Problem 2.49 is a special case with \(a=m \omega / 2 \hbar.\) (d) Find \(K\) for a free particle of mass \(m\). In this case the stationary states are continuous, not discrete, and one must make the replacement \(\sum_{n} \rightarrow \int_{-\infty}^{\infty} d p\) in Equation 6.79. (e) Find \(\Psi(x, t)\) for a free particle that starts out in the state \(\Psi(x, 0)=\left(\frac{2 a}{\pi}\right)^{1 / 4} e^{-a x^{2}}\) Compare your answer with Problem 2.21.
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