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What, exactly, does Schrödinger's wave equation describe (just in plain English, without any of the math please)

No, the wavefunction is not the probability of finding the particle at some location, rather, the probability is something similar to the square of the wavefunction, but a little more complicated. So ...
Pato Galmarini's user avatar
1 vote

What, exactly, does Schrödinger's wave equation describe (just in plain English, without any of the math please)

Unfortunately, in a specific sense your description is quite wrong. Many of the aspects that you describe are interpretations that arose years after the framework was first published. You are ...
Cleonis's user avatar
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What, exactly, does Schrödinger's wave equation describe (just in plain English, without any of the math please)

It's hard to say whether its correct or not...Because its hard to say whether our understanding are the same. For the key part of your statement In the classical equation, Schrodinger replaced the ...
MakiseKurisu's user avatar
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Schroedinger equation applicable only in electron

Protons and neutrons are not elementary particles (they are made up of quarks), while (as far as we know) electrons are. So protons and neutrons are similar to any other object made up of multiple ...
Stefania's user avatar
1 vote
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Calculating the expectation value of the angular momentum operator

How do come up with $$ \frac{1}{\pi}\frac{\hbar}{i} \frac{1}{2\phi}\sin^2{2\phi \pi}? $$ The answer for the integral must be a number, not a function of $\phi$. The integral $$ \int_0^{2 \pi} \cos\...
mike stone's user avatar
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1 vote

Quantum Mechanical Current Normalisation

You just need to normalise your solution to unity, after correcting your expressions. The meaning of this is known as the Born rule.
my2cts's user avatar
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2 votes
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Quantum Mechanical Current Normalisation

Well, the Schroedinger equation for $\psi$ is always kind of the same; the potential energy or the number of kinetic terms changes, but otherwise the form $$ \hat{H}\psi = E\psi $$ remains. But the ...
Ján Lalinský's user avatar
1 vote

Question on 1D Scattering Resonances

In the theory of scattering one defines phase shift $\delta$ as the change in phase of the outgoing to incoming wave. At those particular points where a resonance occurs this quantity is $\pi$/2. The ...
SAKhan's user avatar
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1 vote
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Question on 1D Scattering Resonances

Resonances in general are associated with quasi-bound (metastable) states of the scattering potential. When the energy of the falling wave is close to the energy of the metastable state, the ...
E. Anikin's user avatar
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3 votes

Negative kinetic energy on a step potential

Negative kinetic energy is absurd, right? What's wrong with this calculation? Kinetic energy value we assign to the particle, based on measurement or the psi function, cannot be negative. When you ...
Ján Lalinský's user avatar
1 vote

Relationship between unitaries generated by a Hamiltonian and its negative sign

Short answer: Your guess is correct, for general time-dependent Hamiltonians $U_1^\dagger(t)\neq U_2(t)$ and there is usually no relation between these two (we will give two simple counterexamples ...
Frederik vom Ende's user avatar
7 votes
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Negative kinetic energy on a step potential

I'm having trouble with the explanation of the kinetic energy on the classically forbbiden region on a step potential ($V=0$ for $x<0$, $V=V_0$ for $x>0$ and $E<V_0$). ... On the classically ...
hft's user avatar
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Negative kinetic energy on a step potential

Expectation values obey classical rules (Ehrenfest's theorem) and to be in such a region where the potential energy is positive and higher than your total energy you would need to have a negative ...
mike1994's user avatar
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Time derivative of complex conjugate wave function

Call your first displayed equation $E(\Psi)$ and your second $F(\Psi^*)$. Theorem: Suppose $E(\Psi)$ is true. Then $E(\Psi^*)$ is true if and only if ${\partial \Psi/\partial t}=0$. Proof: If $E(\...
WillO's user avatar
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3 votes
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Time derivative of complex conjugate wave function

Is $\Psi^*$ a wavefunction? Depends on what you mean by that: $\Psi^*(x,t)$ is certainly a function, it has the property that $\int dx\, |\Psi^*|^2 = 1$ for all $t$ (assuming that $\Psi$ does), so you ...
Javier's user avatar
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1 vote

Complex Conjugate of Wave Function's Derivative

Basically it goes like this : $$\begin{align} \left( \frac{i\hbar}{2m} \frac{\partial^2 \Psi}{\partial x^2} - \frac{i}{\hbar}V\Psi \right )^* &=\\ \left( \frac{i\hbar}{2m} \frac{\partial^2 \Psi}{...
Agnius Vasiliauskas's user avatar
2 votes

Derivation of Schrödinger equation in Feynman-Hibbs

Briefly speaking, it follows from dimensional analysis that higher-order terms ${\cal O}(\eta^{n\geq 3})$ will [after the Gaussian $\eta$-integration (4.5)] only produce higher-orders terms ${\cal O}(\...
Qmechanic's user avatar
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5 votes
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Is Schrodinger's cat a problem of how we define identity?

Early versions of quantum theory (called the "Copenhagen Interpretation") contained a (subjectively) weird thing called the "Heisenberg cut" the idea of the Heisenberg cut is (...
ors's user avatar
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3 votes

Is Schrodinger's cat a problem of how we define identity?

The idea that the property of being alive is emergent doesn't mean it is an illusion. There is a real objective difference between a cat being alive and a cat being dead. A living cat breathes in ...
alanf's user avatar
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5 votes

Does this double well potential contradict the fact that there is no degeneracy for one-dimensional bound states?

No, there is no contradiction. For any finite height of the barrier, the splitting between eigenvalues remains small but nonzero, and the result holds. If you truly want to think of the barrier as ...
Emilio Pisanty's user avatar
7 votes
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Does this double well potential contradict the fact that there is no degeneracy for one-dimensional bound states?

If you take $V_0$ infinite, the wave function solution of Schrödinger's equation $\psi(x)$ is forced to vanish on the barrier. So it seems reasonable that the solution on either side of the barrier is ...
Mateo's user avatar
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Is the energy of an orbital dependent on temperature?

Temperaturile foarte inalte peste 1000 de grade pot ioniza atomii fapt binecunoscut de la tuburile electronice, dar nu pot modifica dimensiunile atomilor sau a ionilor respectivi. Anumite radiatii de ...
Virgil Ioan's user avatar
1 vote
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What's the meaning of the momentum operator?

Disperson relation Before even looking at the Schrödinger eqaution let's look at the wave equation to understand an important concept in physics: the dispersion relation. The wave equation (1D) is ...
AccidentalTaylorExpansion's user avatar
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What's the meaning of the momentum operator?

Quantum mechanics uses notions of systems, states, observables, and dynamics. These notions pervade all textbook physical theories. A system is any object we want to study the properties of. We say ...
Silly Goose's user avatar
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7 votes
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Time-evolution operator in QFT

Maggiore is assuming there that the field interacts with itself or with other fields in a stationary way. In other words, there is no direct appearance of time in the total Hamiltonian. This is a ...
Valter Moretti's user avatar
1 vote

I need to find the state of the system at a general time, knowing the Hamiltonian and the state at $t=0$

The Hamiltonian for a certain three-level system is represented by the matrix $$H = \begin{pmatrix}a & 0 & b \\ 0 & c & 0 \\ b & 0 & a\end{pmatrix},$$ where $a$, $b$, and $c$ ...
hft's user avatar
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4 votes

I need to find the state of the system at a general time, knowing the Hamiltonian and the state at $t=0$

For this particular initial state, you do not need to diagonalize the Hamiltonian. The reason is that this particular initial state is an eigenstate of the Hamiltonian, and as a consequence you can ...
Níckolas Alves's user avatar

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