/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 48 Consider the Hamiltonian of a sp... [FREE SOLUTION] | 91Ó°ÊÓ

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Consider the Hamiltonian of a spinless particle of charge \(e\). In the presence of a static magnetic field, the interaction terms can be generated by $$ \mathbf{p}_{\text {operator }} \rightarrow \mathbf{p}_{\text {operator }}-\frac{e \mathbf{A}}{c} $$ where \(\mathbf{A}\) is the appropriate vector potential. Suppose, for simplicity, that the magnetic field \(\mathbf{B}\) is uniform in the positive \(z\)-direction. Prove that the above prescription indeed leads to the correct expression for the interaction of the orbital magnetic moment \((e / 2 m c) \mathbf{L}\) with the magnetic field \(\mathbf{B}\). Show that there is also an extra term proportional to \(B^{2}\left(x^{2}+y^{2}\right)\), and comment briefly on its physical significance.

Short Answer

Expert verified
Replacing \( \mathbf{p} \) produces an orbital interaction and a harmonic oscillator term due to \( B^2(x^2 + y^2) \), representing transverse confinement.

Step by step solution

01

Understand the Problem

We need to show that replacing \( \mathbf{p} \) with \( \mathbf{p} - \frac{e\mathbf{A}}{c} \) in the Hamiltonian generates the correct interaction term for the orbital magnetic moment with a magnetic field \( \mathbf{B} \), and identify an extra term that is proportional to \( B^2(x^2 + y^2) \).
02

Express the Hamiltonian with Magnetic Interaction

The original Hamiltonian for a particle of mass \( m \) and charge \( e \) without a magnetic field is given by \( H = \frac{\mathbf{p}^2}{2m} \). When a magnetic field is present, we replace \( \mathbf{p} \) with \( \mathbf{p} - \frac{e\mathbf{A}}{c} \), so the Hamiltonian becomes: \[ H = \frac{1}{2m} \left( \mathbf{p} - \frac{e\mathbf{A}}{c} \right)^2. \]
03

Simplify the Expressed Hamiltonian

Expand the Hamiltonian using the expression: \[ \left( \mathbf{p} - \frac{e\mathbf{A}}{c} \right)^2 = \mathbf{p}^2 - \frac{e}{c} (\mathbf{p}\cdot\mathbf{A} + \mathbf{A}\cdot\mathbf{p}) + \frac{e^2}{c^2}\mathbf{A}^2. \]Thus, \[ H = \frac{\mathbf{p}^2}{2m} - \frac{e}{2mc} (\mathbf{p}\cdot\mathbf{A} + \mathbf{A}\cdot\mathbf{p}) + \frac{e^2}{2mc^2}\mathbf{A}^2. \]
04

Connect with the Orbital Magnetic Moment

The term \( -\frac{e}{2mc} (\mathbf{p}\cdot\mathbf{A} + \mathbf{A}\cdot\mathbf{p}) \) can be related to the interaction of the orbital magnetic moment \((e/2mc) \mathbf{L} \) with the magnetic field. By taking \( abla \times \mathbf{A} = \mathbf{B} \), where \( \mathbf{L} = \mathbf{r} \times \mathbf{p} \), this term represents the coupling \( (e/2mc) \mathbf{L} \cdot \mathbf{B} \).
05

Identify the Extra Term

In the expanded expression of the Hamiltonian, the term \( \frac{e^2}{2mc^2} \mathbf{A}^2 \) is an additional term. For a uniform \( \mathbf{B} \) field in the \( z \)-direction, the vector potential \( \mathbf{A} \) can often be chosen as \( \mathbf{A} = \frac{1}{2} (-yB, xB, 0) \). The \( \mathbf{A}^2 \) term becomes proportional to \( B^2 (x^2 + y^2) \), which represents a harmonic oscillator potential in the transverse directions.
06

Conclude with the Physical Significance

The additional harmonic oscillator term \( B^2 (x^2 + y^2) \) indicates that the particle experiences a restoring force proportional to its displacement in the \( xy \) plane due to the magnetic field, which can lead to phenomena such as Landau levels in quantum mechanics.

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Key Concepts

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

Hamiltonian
The Hamiltonian in quantum mechanics represents the total energy of a system. In the context of a charged particle in a magnetic field, it accounts for both kinetic and potential energies. Here, for a spinless particle with mass \( m \) and charge \( e \), the typical Hamiltonian without an external field is given by \( H = \frac{\mathbf{p}^2}{2m} \). This represents a purely kinetic scenario where only the momentum \( \mathbf{p} \) and mass matter.

To incorporate the effect of a magnetic field, the Hamiltonian is modified by substituting the momentum \( \mathbf{p} \) with \( \mathbf{p} - \frac{e\mathbf{A}}{c} \). Here, \( \mathbf{A} \) represents the vector potential associated with the magnetic field. This substitution is crucial as it modifies the energy landscape seen by the particle, introducing interaction terms that account for the influence of the field.
  • The term \( -\frac{e}{2mc} (\mathbf{p}\cdot\mathbf{A} + \mathbf{A}\cdot\mathbf{p}) \) indicates how the orbital angular momentum \( \mathbf{L} \) interacts with the magnetic field.
  • The presence of \( \frac{e^2}{2mc^2}\mathbf{A}^2 \) reflects additional energy due to the field configuration.
Ultimately, this Hamiltonian provides a comprehensive framework describing energy due to motion (kinetic) and position (potential due to magnetic interaction).
Orbital Magnetic Moment
The orbital magnetic moment is fundamentally linked to an electron's motion around an atomic nucleus. It is analogous to a tiny loop of current, generating a magnetic field. Denoted by \( \mathbf{\mu} \), the orbital magnetic moment can be expressed as \( (e/2mc) \mathbf{L} \). Here, \( \mathbf{L} = \mathbf{r} \times \mathbf{p} \) represents the orbital angular momentum, where \( \mathbf{r} \) is the position vector, and \( \mathbf{p} \) the momentum.

The interaction of the orbital magnetic moment with an external magnetic field \( \mathbf{B} \) results in an energy term \( \mathbf{\mu} \cdot \mathbf{B} \). This leads to the potential energy associated with the orientation of the magnetic moment in the field.
  • If \( \mathbf{\mu} \) and \( \mathbf{B} \) are aligned, the particle's energy is minimized.
  • If they are opposed, the energy increases.
Thus, this interaction can be seen in atomic and molecular systems where external magnetic fields influence electron orbitals, leading to phenomena like Zeeman splitting in spectral lines.
Vector Potential
The vector potential \( \mathbf{A} \) is a foundational concept used to describe magnetic fields within electrodynamics. It is a vector field with important implications for the magnetic fields it represents. The relationship between the vector potential and the magnetic field \( \mathbf{B} \) is given by \( abla \times \mathbf{A} = \mathbf{B} \), which implies that the magnetic field is the curl of the vector potential.

In quantum mechanics, the vector potential plays a critical role beyond simply describing the field. It directly influences the wave function of charged particles, affecting their dynamics as seen in the Hamiltonian. For a uniform magnetic field in the z-direction, the vector potential is often chosen as \( \mathbf{A} = \frac{1}{2}(-yB, xB, 0) \). This choice simplifies calculations while capturing the essential physics of a uniform magnetic field.
  • It effectively introduces terms within the Hamiltonian such as \( (\mathbf{p}\cdot\mathbf{A} + \mathbf{A}\cdot\mathbf{p}) \) and \( \mathbf{A}^2 \), representing energies due to magnetic interactions.
  • It underpins key quantum phenomena including the Aharonov-Bohm effect, highlighting the nontrivial role of potentials beyond field strengths alone.
Understanding \( \mathbf{A} \) is crucial for deep insights into how magnetic fields impact quantum systems.
Uniform Magnetic Field
A uniform magnetic field is one where the magnetic field lines are parallel and equally spaced throughout the region of interest. This simple structure helps in understanding the effects of magnetic fields on charged particles such as electrons. When the uniform magnetic field is aligned along the positive z-direction, it can significantly simplify analysis in physics problems.

In our exercise, this uniformity results in a consistent influence on the particle regardless of its position in the z-direction, introducing a straightforward yet rich interaction landscape.
  • The vector potential \( \mathbf{A} \) is often chosen smartly to match the uniform condition, leading to simple forms like \( \mathbf{A} = \frac{1}{2}(-yB, xB, 0) \).
  • Key effects such as the emergence of a harmonic oscillator-like potential, described by terms \( B^2 (x^2 + y^2) \), become evident.
This additional term implies that within the magnetic field, there is an induced restoring force proportional to the displacement from the origin in the xy-plane. This setup gives rise to phenomena like Landau quantization, where energy levels are quantized due to the magnetic field's input. Thus, a uniform magnetic field creates a predictable yet fascinating context for exploring magnetic interactions in quantum systems.

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Most popular questions from this chapter

A particle with mass \(m\) moves in one dimension and is acted on by a constant force \(F\). Find the operators \(x(t)\) and \(p(t)\) in the Heisenberg picture, and find their expectation values for an arbitrary state \(|\alpha\rangle\). Use \(\langle x(0)\rangle=x_{0}\) and \(\langle p(0)\rangle=p_{0}\). The result should be obvious. Comment on how to do this problem in the Schrödinger picture, but do not try to work it through.

Consider a particle moving in one dimension under the influence of a potential \(V(x)\). Suppose its wave function can be written as \(\exp [i S(x, t) / \hbar]\). Prove that \(S(x, t)\) satisfies the classical Hamilton-Jacobi equation to the extent that \(\hbar\) can be regarded as small in some sense. Show how one may obtain the correct wave function for a plane wave by starting with the solution of the classical Hamilton-Jacobi equation with \(V(x)\) set equal to zero. Why do we get the exact wave function in this particular case?

Make the definitions $$ J_{\pm} \equiv \hbar a_{\pm}^{\dagger} a_{\mp}, \quad J_{z} \equiv \frac{\hbar}{2}\left(a_{+}^{\dagger} a_{+}-a_{-}^{\dagger} a_{-}\right), \quad N \equiv a_{+}^{\dagger} a_{+}+a_{-}^{\dagger} a_{-} $$ where \(a_{\pm}\)and \(a_{\pm}^{\dagger}\) are the annihilation and creation operators of two independent simple harmonic oscillators satisfying the usual simple harmonic oscillator commutation relations. Also make the definition $$ \mathbf{J}^{2} \equiv J_{z}^{2}+\frac{1}{2}\left(J_{+} J_{-}+J_{-} J_{+}\right) $$ Prove $$ \left[J_{z}, J_{\pm}\right]=\pm \hbar J_{\pm}, \quad\left[\mathbf{J}^{2}, J_{z}\right]=0, \quad \mathbf{J}^{2}=\left(\frac{\hbar^{2}}{2}\right) N\left[\left(\frac{N}{2}\right)+1\right] $$

Consider a one-dimensional simple harmonic oscillator. a. Using $$ \left.\left.\begin{array}{l} a \\ a^{\dagger} \end{array}\right\\}=\sqrt{\frac{m \omega}{2 \hbar}}\left(x \pm \frac{i p}{m \omega}\right), \quad \begin{array}{l} a|n\rangle \\ a^{\dagger}|n\rangle \end{array}\right\\}=\left\\{\begin{array}{l} \sqrt{n}|n-1\rangle \\ \sqrt{n+1}|n+1\rangle \end{array}\right. $$ evaluate \(\langle m|x| n\rangle,\langle m|p| n\rangle,\langle m|\\{x, p\\}| n\rangle,\left\langle m\left|x^{2}\right| n\right\rangle\), and \(\left\langle m\left|p^{2}\right| n\right\rangle .\) b. Translated from classical physics, the virial theorem states that $$ \left\langle\frac{\mathbf{p}^{2}}{m}\right\rangle=\langle\mathbf{x} \cdot \nabla V\rangle \text { (3D) } \quad \text { or } \quad\left\langle\frac{p^{2}}{m}\right\rangle=\left\langle x \frac{d V}{d x}\right\rangle $$ Check that the virial theorem holds for the expectation values of the kinetic and the potential energy taken with respect to an energy eigenstate.

Consider a particle of mass \(m\) subject to a one-dimensional potential of the following form: $$ V=\left\\{\begin{array}{lll} \frac{1}{2} k x^{2} & \text { for } & x>0 \\ \infty & \text { for } & x<0 \end{array}\right. $$ a. What is the ground-state energy? b. What is the expectation value \(\left\langle x^{2}\right\rangle\) for the ground state?

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