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Quantum mechanics describes the microscopic properties of nature in a regime where classical mechanics no longer applies. It explains phenomena such as the wave-particle duality, quantization of energy, and the uncertainty principle and is generally used in single-body systems. Use the quantum-field-theory tag for the theory of many-body quantum-mechanical systems.
5
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Eigenvalue spectrum of $L_x+iL_y$
Is it possible to find out the generic eigenvalue spectrum of the non-Hermitian operator $L_x+iL_y$, without using any representation?
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1
answer
528
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Sources of zero point energy in quantum mechanics and free quantum field theory
A quantum linear harmonic oscillator has a definite non-zero ground state energy $E_0=\frac{1}{2}\hbar\omega\neq 0$. However, in this energy eigenstate, the position and momenta are uncertain and thei …
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1
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A question from Quantum Scattering theory Sakurai
This is a question from scattering theory - solution for incoming plane wave and outgoing scattered wave:
In Sakurai's Modern Quantum Mechanics, I encountered the following line:
Furthermore, bec …
4
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2
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What determines the Hilbert space of quantum mechanical system?
Is there a systematic way to determine the Hilbert space of a given quantum mechanical problem? Given a problem, the first task is to determine the Hilbert space. Is it dictated by the boundary condit …
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4
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An apparent contradiction with basis transformation in quantum mechanics
Under a change of basis i.e., transforming from one orthonormal $\{|\phi_n\rangle\}$ base to another $\{|\phi^\prime_n\rangle\}$ (when looked from a passive point of view) implies that the state doesn …
1
vote
Accepted
Confusion on quantum numbers
Hydrogen atom with and without Coulomb potential
For H-atom with Coulomb potential and no perturbations such as spin-orbit interaction, relativistic correction etc, $n,l,m_l,m_s$ are good (conserved) …
3
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1
answer
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Any other bound state problems using $a_+$ and $a_-$?
Why is it that creation and annihilation operators ($a_+$ and $a_-$) can only be defined for the problem of quantum harmonic oscillator and nothing else? Can any other bound state problem be solves us …
10
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1
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Formalism Of Quantum Field Theory vs Quantum Mechanics
How far can we extend the formalisms on quantum mechanics (QM) to quantum field theory (QFT)? In particular,
How is a Fock space $\mathcal{F}$ different from a Hilbert space $\mathcal{H}$? Can a gen …
3
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The quantum state just after a position measurement
When the position is measured on this state, the wavefunction collapses to a position eigenstate which is a Dirac-delta function $\delta(x-a)$ spiked about $x=a$. The measurement of position on any wa …
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3
answers
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Why should the perturbation be small and in what sense?
In time-independent perturbation theory, one writes $$\hat{H}=\hat{H}_0+\lambda \hat{H}^\prime$$ where $\lambda H^\prime$ is a "small" perturbation.
Why should the perturbation be small for perturb …
2
votes
1
answer
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What underlies super-selection rule in quantum field theory?
Super-selection rule in quantum field theory states that superpositions of two states with different charges do no exist in nature.
What does "charge" mean in this context?
Is there a deeper princi …
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2
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Is there a number operator for a non-degenerate two-level system?
The Hamiltonian of a two-level system is given by $$H=E_1|1\rangle\langle 1|+E_2|2\rangle\langle 2|$$ where both the energy eigenstates $|1\rangle$ and $|2\rangle$ are non-degenerate with $E_2>E_1$. N …
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How is a macrostate specified in quantum statistics?
It is easy to understand the idea of macrostate and microstate in the context of classical statistical mechanics. The description of a microstate of a system requires the specification of position and …
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3
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How to see that $[\textbf{p}^2,\textbf{L}^2]=[\textbf{p}^4,\textbf{L}^2]=0$ without doing an...
The Hamiltonian of a particle moving under the influence of a central potential $V(r)$ given by $$H=\frac{\textbf{p}^2}{2m}+V(r)$$ commutes with $\textbf{L}^2\equiv L_x^2+L_y^2+L_z^2$. Without doing a …
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Sufficient condition for square integrability [duplicate]
The necessary condition for $\int\limits_{-\infty}^{+\infty}|\psi(x)|^2dx$ to be integrable is that $\psi(x)\rightarrow 0$ as $x\rightarrow\pm\infty$. But this is not the sufficient condition. For ex …