Explanations 
Textbooks 
Exams 
GCSE 
IGCSE 
AS 
A Level 
International A Level 
University Admissions Tests 
Flashcards 
Vaia AI 
Notes 
Study Plans 
Study Sets 
Find a job 
Find a degree 
Student Deals 
Magazine 
Mobile App Chapter 7: Problem 9
If \(H=p^{2} / 2 \mu+V(x)\) for a one-dimensional particle, and \(V(x)\) can be expressed as a power series in \(x\), show by purely matrix methods that $$ \frac{d x}{d t}=\frac{p}{\mu}, \quad \frac{d p}{d t}=-\frac{d V}{d x} $$ What is meant by the operator \(d / d t\) in this connection?
Short Answer
Expert verified
\(\frac{dx}{dt} = \frac{p}{\mu}\), \(\frac{dp}{dt} = -\frac{dV}{dx}\). The operator \(\frac{d}{dt}\) indicates the change over time in the Heisenberg picture.
Step by step solution
01
Express Hamiltonian in Matrix Form
The Hamiltonian for the one-dimensional particle is given by \[ H = \frac{p^{2}}{2 \mu} + V(x). \]Express it as a matrix operator acting on a state vector \[ H = T + V, \]where \[ T = \frac{p^{2}}{2 \mu} \] is the kinetic energy operator and \( V \) is the potential energy operator.
02
Derive Equations of Motion
Using the Heisenberg equation of motion, we have:\[ \frac{dA}{dt} = \frac{i}{\hbar} [H, A] + \frac{\partial A}{\partial t} \]For the position operator \( x \):\[ \frac{dx}{dt} = \frac{i}{\hbar} [H, x] + \frac{\partial x}{\partial t}. \]Since \( x \) is not an explicit function of \( t \), \(\frac{\partial x}{\partial t} = 0\). Thus,\[ \frac{dx}{dt} = \frac{i}{\hbar} [H, x]. \]
03
Evaluate Commutator for Position
Compute the commutator \( [H, x] \):\[ [H, x] = \left[ \frac{p^2}{2\mu} + V(x), x \right]. \]Since \( [f(x), x] = 0 \), the second term vanishes and we are left with:\[ \left[ \frac{p^2}{2\mu}, x \right]. \]Using the commutation relation \( [p, x] = -i \hbar \),\[ \left[ p^2, x \right] = p [p, x] + [p, x] p = p(-i \hbar) + (-i \hbar) p = -2i \hbar p. \]Thus,\[ \left[ \frac{p^2}{2\mu}, x \right] = \frac{-2i \hbar p}{2\mu} = \frac{-i \hbar p}{\mu}. \]Hence,\[ \frac{dx}{dt} = \frac{i}{\hbar} \cdot \frac{-i \hbar p}{\mu} = \frac{p}{\mu}. \]
04
Evaluate Commutator for Momentum
Now consider the momentum operator \( p \):\[ \frac{dp}{dt} = \frac{i}{\hbar} [H, p] + \frac{\partial p}{\partial t}. \]Since \( p \) is not an explicit function of \( t \), we have:\[ \frac{dp}{dt} = \frac{i}{\hbar} [H, p]. \]Compute the commutator \( [H, p] \):\[ [H, p] = \left[ \frac{p^2}{2\mu} + V(x), p \right]. \]The first term vanishes and we are left with:\[ [V(x), p]. \]Using the commutation relation \( [x, p] = i \hbar \),\[ [V(x), p] = [V(x), x] [x, p] = V'(x) \cdot (i \hbar) = -i \hbar V'(x). \]Therefore,\[ \frac{dp}{dt} = \frac{i}{\hbar} \cdot (-i \hbar) V'(x) = -\frac{dV}{dx}. \]
05
Define Operator \(d/dt\)
In this context, the operator \( \frac{d}{dt} \) refers to the time evolution of an operator in the Heisenberg picture. It denotes the rate of change of the operator with respect to time using the commutation relations with the Hamiltonian.
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with Vaia!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Heisenberg Equation of Motion
The Heisenberg equation of motion is a foundational concept in quantum mechanics that tells us how operators evolve over time in the Heisenberg picture. This approach is particularly useful for studying the dynamics of a system without solving the Schrödinger equation directly.
The Heisenberg equation is given by: \[ \frac{dA}{dt} = \frac{i}{\hbar} [H, A] + \frac{\partial A}{\partial t} \]
Here, \( A \) represents an operator such as position \(x\) or momentum \(p\), and \(H\) is the Hamiltonian of the system. The term \( \frac{i}{\hbar} [H, A] \) represents the commutator of the Hamiltonian and the operator, while \( \frac{\partial A}{\partial t} \) accounts for any explicit time dependence of the operator.
For operators that do not explicitly depend on time, the equation simplifies to:
- \( \frac{dA}{dt} = \frac{i}{\hbar} [H, A] \)
In this exercise, we used the Heisenberg equation to derive the time evolution of the position and momentum operators, showing that:
- \( \frac{dx}{dt} = \frac{p}{\mu} \)
- \( \frac{dp}{dt} = -\frac{dV}{dx} \)
These equations match the classical equations of motion derived from Newton's laws.
The Heisenberg equation is given by: \[ \frac{dA}{dt} = \frac{i}{\hbar} [H, A] + \frac{\partial A}{\partial t} \]
Here, \( A \) represents an operator such as position \(x\) or momentum \(p\), and \(H\) is the Hamiltonian of the system. The term \( \frac{i}{\hbar} [H, A] \) represents the commutator of the Hamiltonian and the operator, while \( \frac{\partial A}{\partial t} \) accounts for any explicit time dependence of the operator.
For operators that do not explicitly depend on time, the equation simplifies to:
- \( \frac{dA}{dt} = \frac{i}{\hbar} [H, A] \)
In this exercise, we used the Heisenberg equation to derive the time evolution of the position and momentum operators, showing that:
- \( \frac{dx}{dt} = \frac{p}{\mu} \)
- \( \frac{dp}{dt} = -\frac{dV}{dx} \)
These equations match the classical equations of motion derived from Newton's laws.
Commutator
Understanding commutators is crucial in quantum mechanics. A commutator of two operators \(A\) and \(B\) is defined as: \[ [A, B] = AB - BA \]
This tells us how much two operations differ when applied in sequence. In the context of our exercise, the commutators help us determine the rate of change of operators like position and momentum.
For instance, the position-momentum commutation relation is:
\( [x, p] = i \hbar \)
This fundamental commutation relation tells us that position and momentum cannot be simultaneously known to arbitrary precision—a cornerstone of the Heisenberg uncertainty principle.
In the exercise:
- We computed \( [H, x] \) to find \( \frac{dx}{dt} \), showing that \( \left[ \frac{p^2}{2\mu}, x \right] = \frac{-i \hbar p}{\mu} \), leading to \( \frac{dx}{dt} = \frac{p}{\mu} \).
- We also computed \( [H, p] \) to find \( \frac{dp}{dt} \), showing that \( [V(x), p] = -i \hbar \frac{dV}{dx} \), leading to \( \frac{dp}{dt} = -\frac{dV}{dx} \).
The commutator plays a key role in these derivations, linking the quantum description with classical dynamics.
This tells us how much two operations differ when applied in sequence. In the context of our exercise, the commutators help us determine the rate of change of operators like position and momentum.
For instance, the position-momentum commutation relation is:
\( [x, p] = i \hbar \)
This fundamental commutation relation tells us that position and momentum cannot be simultaneously known to arbitrary precision—a cornerstone of the Heisenberg uncertainty principle.
In the exercise:
- We computed \( [H, x] \) to find \( \frac{dx}{dt} \), showing that \( \left[ \frac{p^2}{2\mu}, x \right] = \frac{-i \hbar p}{\mu} \), leading to \( \frac{dx}{dt} = \frac{p}{\mu} \).
- We also computed \( [H, p] \) to find \( \frac{dp}{dt} \), showing that \( [V(x), p] = -i \hbar \frac{dV}{dx} \), leading to \( \frac{dp}{dt} = -\frac{dV}{dx} \).
The commutator plays a key role in these derivations, linking the quantum description with classical dynamics.
Position and Momentum Operators
Position \((x)\) and momentum \((p)\) operators are core to understanding quantum mechanics. These operators are not just numbers but mathematical objects that act on quantum states.
The position operator \( x \) represents the position of a particle in space, while the momentum operator \( p \) represents the particle's momentum. In quantum mechanics:
- The position operator \( x \) acts multiplicatively on a wavefunction \( \psi(x) \), meaning it simply multiplies the wavefunction by \( x \).
- The momentum operator \( p \) is represented as \( -i \hbar \frac{\partial}{\partial x} \) when acting on a wavefunction in the position representation.
These operators do not commute, as shown by the fundamental relation:
\( [x, p] = i \hbar \)
This non-commutativity is a direct expression of the uncertainty principle.
For the problem at hand, we used the matrix methods to derive:
- The time evolution of the position operator: \( \frac{dx}{dt} = \frac{p}{\mu} \)
- The time evolution of the momentum operator: \( \frac{dp}{dt} = -\frac{dV}{dx} \)
These expressions connect quantum mechanical operators to classical quantities, bridging quantum mechanics with classical mechanics.
The position operator \( x \) represents the position of a particle in space, while the momentum operator \( p \) represents the particle's momentum. In quantum mechanics:
- The position operator \( x \) acts multiplicatively on a wavefunction \( \psi(x) \), meaning it simply multiplies the wavefunction by \( x \).
- The momentum operator \( p \) is represented as \( -i \hbar \frac{\partial}{\partial x} \) when acting on a wavefunction in the position representation.
These operators do not commute, as shown by the fundamental relation:
\( [x, p] = i \hbar \)
This non-commutativity is a direct expression of the uncertainty principle.
For the problem at hand, we used the matrix methods to derive:
- The time evolution of the position operator: \( \frac{dx}{dt} = \frac{p}{\mu} \)
- The time evolution of the momentum operator: \( \frac{dp}{dt} = -\frac{dV}{dx} \)
These expressions connect quantum mechanical operators to classical quantities, bridging quantum mechanics with classical mechanics.
