Linked Questions

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Galilean transformation of Schrodinger equation and momentum operator [duplicate]

Let $$ \left.\begin{aligned} t'&=t\\x'&=x-vt \end{aligned}\right\} \quad \Longrightarrow\quad \dot{x}'=\dot{x}-v $$ and therefore $p'=p-mv$. If $p'=-i\hbar\nabla' $, then $\nabla'=\nabla-iv/\...
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0answers
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Galilean invariance of the free schroedinger equation [duplicate]

My question follows this question: Naive interpretation of Galilean invariance of the TDSE Essentially, I'm not sure how to proceed mathematically. We have the transformations: $$\begin{cases}x'=x-...
123
votes
2answers
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Reading the Feynman lectures in 2012

The Feynman lectures are universally admired, it seems, but also a half-century old. Taking them as a source for self-study, what compensation for their age, if any, should today's reader undertake? ...
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1answer
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Boosts are non-unitary!

Unlike rotations, the boost transformations are non-unitary. Therefore, the boost generators are not Hermitian. When boosts induce transformations in the Hilbert space, will those transformation be ...
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1answer
3k views

What is the wavefunction of the Young Double Slit experiment?

I have never seen the wavefunction for this experiment and would like to know how to derive it using the Schrodinger equation. I specifically want to see how the electron wave function leaves the ...
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2answers
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Why the Galileo transformation are written like this in Quantum Mechanics?

In Quantum Mechanics it is said that the Galileo transformation $$\mathbf{r}\mapsto \mathbf{r}-\mathbf{v}t\quad \text{and}\quad \mathbf{p}\mapsto \mathbf{p}-m\mathbf{v}\tag{1}$$ is given by the ...
9
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4answers
552 views

Uniqueness of the probability function for the Schrödinger equation

David Bohm in Section (4.5) of his wonderful monograph Quantum Theory after defining the usual density probability function $P(x,t)=\psi^{*} \psi$ for the Schrödinger equation for the free particle in ...
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1answer
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Naive interpretation of Galilean invariance of the TDSE

I was told today by someone smarter than myself that the time-dependent Schroedinger equation in one dimension was invariant under a Galilean transformation of $(x,t)$, namely under $$\begin{cases}x'=...
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1answer
282 views

How does one find the wavefunction of a particle in its rest frame?

In classical mechanics, the orbital angular momentum of a particle is defined as $\textbf{L}=\textbf{r}\times\textbf{p}$. This is zero in the rest frame of the particle where $\textbf{p}=0$. Quantum ...
2
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1answer
967 views

Generator of Velocity Transformations - Galilean Transformations

Under a Galilean transformation, the coordinates and momenta of any system transform as: $$ t \rightarrow t',\\ \vec r\:' = \vec r + \vec vt,\\ \vec p\:' = \vec p + m\vec v $$ where $\vec v$ is ...
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1answer
452 views

Derive the Lagrangian that yields the free Schrödinger's equation from Galileian Invariance

The Lagrangian Density $$L(\Psi, \Psi^*)=i \hbar \dot{\Psi} \Psi^* + \frac{\hbar^2}{2m} \Psi \Delta \Psi^*$$ will yield the schroedinger equations for $\Psi$ and $\Psi^*$. Can we derive this ...
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1answer
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Galilean relativity in QM

Intro I've been trying to show that the generator of boosts can be written in operator form as can be seen here, as: $$ B = \sum_i m_i x_i(t) - t \sum_i p_i $$ As a reminder the transformation ...
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1answer
148 views

Show Galilean invariance of Schrödinger eq

I'm trying to show the Galilean invariance of the (time-dependent) Schrödinger equation by transforming as follows: $$ \left\{\begin{eqnarray}\psi(\vec{r},t) &=& \psi(\vec{r}'-\vec{v}t,t),\\ \...
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0answers
28 views

Zero momentum in Non relativistic Quantum Mechanics and about Dirac matrices

In relativistic quantum mechanics, we can solve the Dirac's equation with an added condition that the momentum of the particle is $0$. However, such independence isn't provided by the Schrodinger's ...