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13 votes

Is the zero vector necessary to do quantum mechanics?

Quantum theory can be completely stated in the complex projective space of a complex Hilbert space. The basic object from this perspective is the transition probability of a couple of rays (pure ...
Valter Moretti's user avatar
10 votes

Quantum: why linear combination of vectors (superposition) is described as "both at the same time"?

“Both at the same time” is actually bad language, especially as this is a basis-dependent statement. For instance, the eigenstate $\vert \uparrow\rangle_x$ of $\sigma_x$ is one state. If you make a ...
ZeroTheHero's user avatar
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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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5 votes
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Why we use trace-class operators and bounded operators in quantum mechanics?

I think Tobias Fünke has essentially answerered this question already, but to be as explicit as possible: we need the states to be trace-class so we can obtain normalized probabilities (just as in the ...
ors'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
4 votes
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Some confusion about understanding the relativistic quantum mechanics

The group is still the Poincare group (Lorentz+ translations). The tricky thing is that we need to find a way for that group to act on the Rays in such a way that it preserves the probability. There ...
Josh Newey's user avatar
4 votes
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Problem Deriving "The General Uncertainty Principle" in Section 5.7 of Susskind's "Quantum Mechanics"

You are correct, but there's one last simplifying step. The norm of the product is the product of the norms. For two complex numbers, $z_1$ and $z_2$: \begin{equation} |z_1\cdot z_2| = |z_1|\cdot|z_2| ...
Aiden's user avatar
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4 votes

Operator's definition in Dirac picture

How did you get from the 1st equation to the second equation? The Second equation will be a number, not an operator. You can't just move the ket to the right. The top equation is an outer product but ...
Josh Newey's user avatar
3 votes
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Can the Parity Operator in polar coordinates be defined as $\hat\Pi\psi(r,\theta,\phi) = \psi(r,\theta+\pi,\phi).$?

Take a look at the derivation from which $Y_\ell^m(\theta,\phi)$ is obtained, in Section 4.1.2 of the book (I have the third edition). You should notice that "$m$ is an integer" arises from ...
hendlim's user avatar
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2 votes
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Do different bases of Fock space commute?

For a given decomposition of the Fock space into $n$-particle subspaces, creation operator commute. For a $1$ particle state $|\psi\rangle$, there is an associated annihilation operator $a(\psi)$. If $...
SolubleFish's user avatar
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2 votes

Is the zero vector necessary to do quantum mechanics?

This is largely a philosophical question, but I would argue that yes, the zero vector is necessary for QM to make sense conceptually. A few examples (off the top of my head) of places where the zero ...
tparker's user avatar
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2 votes

Can I use any linearly independent, orthogonal, eigenkets as starting basis to construct $S_x$, $S_y$ and $S_z$?

Of course, you can write the spin operators in the basis you prefer. If you choose to work in the basis of the eigenstates of $S_z$ $\{|\uparrow\rangle,|\downarrow\rangle\}$, you get $S_x = \frac{\...
Matteo's user avatar
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2 votes
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Alternative way to compute expectation value of momentum?

Yes, you can work out the expectation value of an operator in that way. The key thing to notice is that the expectation value does not depend on which basis you choose to work with. For example, it ...
Níckolas Alves's user avatar
2 votes

How can rotation about Z be a superposition of rotations in X ($|\uparrow \rangle_z = |\uparrow\rangle_x + |\downarrow\rangle _x $)

(I’m sorry to say) there is so much confusion in the title and the post… First, a rotation about $\hat z$ is not a superposition of rotations about $\hat x$. In fact, what you have written has ...
ZeroTheHero's user avatar
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2 votes

How can rotation about Z be a superposition of rotations in X ($|\uparrow \rangle_z = |\uparrow\rangle_x + |\downarrow\rangle _x $)

Think about spin as an internal degree of freedom, similar to the polarization of an electromagnetic wave. This is the approach taken to explain the Stern-Gerlach experiment at the start of Sakurai's &...
agaminon's user avatar
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2 votes

Why we use trace-class operators and bounded operators in quantum mechanics?

The effects are not Hilbert Schmidt in general, so to identify them with HS operators would be too restrictive. However, even if density operators are just a part of the Hilbert-Schmidt operators, the ...
Valter Moretti's user avatar
2 votes
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The eigenvectors associated to the continuous spectrum in Dirac formalism

Dirac is dead, we cannot ask him what he really meant by this. However, the pretension that these eigenstates exist in some sense is prevalent in many texts on quantum mechanics. There are several ...
ACuriousMind's user avatar
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2 votes

The eigenvectors associated to the continuous spectrum in Dirac formalism

As long as you allow elements of the associated rigged Hilbert space the answer is "yes."
mike stone's user avatar
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2 votes
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The meaning of a representation in one-dimensional quantum mechanics

In abstract terms the meaning of representation in these cases is the ordinary meaning of representation in the sense of representation theory. Concretely, you are representing the Heisenberg algebra ...
ACuriousMind's user avatar
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2 votes
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What is the dimension considered in the Schmidt Decomposition?

It is the complex dimension. So for an orthonomal basis of $\mathbb{C}^2$, you would choose two vectors which need not be real. Actually this is the case not just for the Schmidt decomposition but ...
Mateo's user avatar
  • 426
1 vote

How do operators on kets and wavefunctions correspond?

Is it by showing that this holds for position and momentum operators, and then have the result follow from any observable having to be a function of these? Yes, naturally. Your text or instructor ...
Cosmas Zachos's user avatar
1 vote

Operator's definition in Dirac picture

I have a question about the definition of quantum operators in the Dirac picture. The definition is: $$A=\sum_i \sum_j \vert i \rangle A_{ij} \langle j \vert.\tag{1}$$ By deplacing the ket vector I ...
hft's user avatar
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1 vote

Quantum: why linear combination of vectors (superposition) is described as "both at the same time"?

Suppose you're travelling with a bearing of 45 degrees to North. We call this direction "Northeast". Are you travelling North? Some might say yes. Are you travelling East? Some might say yes....
Jagerber48's user avatar
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1 vote

Quantum: why linear combination of vectors (superposition) is described as "both at the same time"?

In quantum theory the evolution of a measurable quantity is described by a linear operator called and observable. The eigenvalues of that observable represent the possible outcomes of a measurement of ...
alanf's user avatar
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1 vote
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Why is the state of a quantum system called "Density $\textbf{Operator}$"?

Respectfully, this is actually a deeper question than it may seem at first. From a purely mathematical standpoint, the answer is pretty trivial: we call the density operator an "operator" ...
tparker's user avatar
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1 vote

Unitarily Inequivalent Representations

For a succinct explanation of the orthogonality of the vaccua for the van Hove model (which seems to be what the initial pdf might have been referring to), see Section 2 of this paper. Tne answer by ...
devnull's user avatar
  • 11
1 vote

Do different bases of Fock space commute?

It is evident linear algebra. Intuition? Try a space of just two bosonic oscillators, i=1,2, $$ [a_i,a^\dagger_j]=\delta_{ij}, \qquad [a_i,a_j]=0, $$ and ditto for the conjugates. Rotate them to $$ \...
Cosmas Zachos's user avatar
1 vote

Bogoliubov transformation of Bunch-Davies vacuum

Yes, that's exactly the point. In $H_\alpha$ we have that $|\alpha\rangle$ is a vacuum state, but in $H_0$ it is a multiparticle (thermal) state. In the same way, in $H_0$ we have that $|0\rangle$ is ...
LolloBoldo's user avatar
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