All Questions
22 questions
0
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0
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123
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Why is psi square a possibility? [duplicate]
Is psi square just an assumption? Or there is a physical reason why they defined like that? My procedure is:
It is intuitive for me to think possibility is proportional to energy distribution. ...
김건형
4
votes
2
answers
104
views
Why is a wave function $\psi$ needed for QM? Is it possible to make a differential equation involving just the p.d.f. $|\psi|^2$ of a particle? [duplicate]
Why do you need a wave function $\psi$ for quantum mechanics? Can't you just make a differential equation involving just the p.d.f. $|\psi|^2$ of a particle? Since basically with quantum mechanics the ...
- 2,180
0
votes
0
answers
59
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Why is the formula for expected value of nonobservables in quantum mechanics different then in regular statistics? [duplicate]
Specifically, why is the operator “sandwiched in” between $\Psi^*$ and $\Psi$? i.e. Why isn’t the formula just $$\langle \hat{Q} \rangle = \int \hat{Q}\cdot|\Psi|^2 dx = \int \hat{Q}\cdot\Psi \cdot \...
- 2,373
0
votes
2
answers
229
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Why particularly probability density is defined as $|\Psi|^2=\Psi \Psi^{*}$?
It may be a stupid question, but why particularly for probability density expression $k~|\Psi|^2 = k~\Psi^{*}\Psi$, it's assumed that $k=1$?
As it is now, then in a complex plane probability density ...
- 14.9k
1
vote
0
answers
32
views
Can I interpret the squaring of the wave function like this? [duplicate]
Born rule states that the probability density of the wave function is equal to the square of the function over the given interval. I thought, "Why squared?". I came up with this:
"We ...
- 786
0
votes
1
answer
880
views
Continuity Equation in Quantum Mechanics when potential is a complex valued function
How can we derive the continuity equation from Schrodinger equation if the potential is a complex function of position?
What I tried was the general $1-D$ derivation of the Continuity equation from ...
- 132
1
vote
2
answers
193
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Question regarding step potential
We are learning about step potential in class. I have completely understood that the behavior of the wave function representing the particle, can have different responses depending on the energy of ...
- 185
3
votes
2
answers
291
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Why do we describe probability amplitude rather than probability itself in quantum mechanics?
In the quantum mechanics, the dynamics of quantum system are described in terms of probability amplitude. However, we want to calculate the probability in the end which can be measured. Why don't we ...
- 479
4
votes
3
answers
327
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Interpretation of the magnetic potential ($A$-field) in the quantum mechanical probability of current
The probability of current in quantum mechanics when the is a magnetic potential, A, is defined as:
$$\boldsymbol j=\frac{1}{2m}(\psi^*\hat{\boldsymbol p} \psi-\psi\hat{\boldsymbol p}\psi^* -2q{\...
- 413
18
votes
3
answers
776
views
Nonexistence of a Probability for Real Wave Equations
David Bohm in Section (4.5) of his wonderful monograph Quantum Theory gives an argument to show that in order to build a physically meaningful theory of quantum phenomena, the wave function $\psi$ ...
- 464
17
votes
3
answers
4k
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Form of Schrödinger equation for the probability density
Is it possible to formulate the Schrödinger equation (SE) in terms of a differential equation involving only the probability density instead of the wave function? If not, why not?
We can take the ...
- 831
0
votes
1
answer
213
views
How to calculate the probability of Hamiltonian Operators being in a certain state?
So, I'm reading through my Quantum Mechanics textbook and I stumbled upon a bit of maths that I'm not entire sure how they got:
\begin{align*}
\mathcal{P}_{a_1} &= \left| e^{-iE_1t/\hbar} \...
- 57
2
votes
0
answers
3k
views
Physical interpretation of a complex potential for a particle in quantum mechanics
In Griffiths' Quantum Mechanics, it is mentioned in a problem that
For an unstable particle that spontaneously disintegrates with a lifetime $\tau$, the total probability of finding the particle ...
- 4,984
22
votes
4
answers
9k
views
Why does the expectation value of an operator $A$ take the form $\langle A\,\rangle=\int{\psi^* (x) A(x) \psi (x) dx}$ in QM?
The following is a quote from an answer I was given to this previous question of mine:
The definition of the expectation value of an operator $A$ is
$$\begin{equation}
\langle A\,\rangle=\int{\...
- 2,520
1
vote
1
answer
877
views
Can the quantum mechanical current density be imaginary?
I am dealing with a situation where I get an imaginary transmission current density.
Is this possible?
Does it imply a zero transmission probability?
- 4,637
-1
votes
2
answers
610
views
What do "ℜe" and "A*" mean?
What do "$\mathfrak{Re}$" and "A*" mean in the following equation (taken from James Binney and David Skinner's QM lecture notes, equation 1.12),
\begin{align}
p(S\text{ or }T)&=\left|A\left(S\text{...
- 387
5
votes
2
answers
8k
views
Why does the magnitude squared of the wave function give us the probability density? [duplicate]
My question doesn't go much beyond the title: Why does $$\left | \psi \left ( x,t \right ) \right |^{2}$$ give us the probability density of something appearing at a certain location? I understand ...
- 90
3
votes
1
answer
247
views
Why do we use $\psi$ instead of a straightforward probability?
What is the advantage/purpose of using $\psi$ for wavefunctions and getting the probability with $|\psi|^2$ as opposed to just defining and using the probability function?
- 1,927
15
votes
3
answers
5k
views
Where does the Born rule come from? [duplicate]
As far as I've read online, there isn't a good explanation for the Born Rule. Is this the case? Why does taking the square of the wave function give you the Probability? Naturally it removes negatives ...
user24082
6
votes
2
answers
2k
views
Amplitude of Probability amplitude. Which one is it?
QM begins with a Born's rule which states that probability $P$ is equal to a modulus square of probability amplitude $\psi$:
$$P = \left|\psi\right|^2.$$
If I write down a wave function like this $\...
- 2,585
2
votes
0
answers
432
views
Probability and probability amplitude [duplicate]
What made scientists believe that we should calculate probability $P$ as the $P = \left|\psi\right|^2$ in quantum mechanics? Was it the double slit experiment? How? Is there anywhere in the ...
- 2,585
7
votes
7
answers
3k
views
Is there a direct physical interpretation for the complex wavefunction?
The Schrödinger equation in non-relativistic quantum mechanics yields the time-evolution of the so-called wavefunction corresponding to the system concerned under the action of the associated ...
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