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As showed by Solovay here, in a non-separable Hilbert space $H$ there may be probability measures that cannot be written, for any $M$ closed subspace of $H$, as $\mu (M)=\mathrm{Tr}[\rho \mathbb{1}_M]$, for some positive self-adjoint trace class $\rho$ with trace 1 (density matrix). Here $\mathbb{1}_M$ denotes the orthogonal projection on $M$. [The proof of ...

2

I usually see it in the reverse way, but it is a matter of taste. Hilbert spaces, in general, can have bases of arbitrarily high cardinality. The specific one used on QM is, by construction, isomorphic to the space L2, the space of square-integrable functions. From there you can show that this particular Hilbert space is separable, because it is a theorem ...

2

State vectors define pure quantum states of a system, and, for an isolated system, evolve with time following the Schrödinger equation (in the Schrödinger picture; in the Heisenberg picture they are constant. Density matrices define classical statistical mixtures of pure quantum states. These can arise in two ways: When we have incomplete information ...

2

Density matrix is NECESSARY when the quantum system is not in a PURE state, i.e. it cannot be described by a wave-function. Such a case is when a beam of freely moving electrons is not spin-polarized, i.e. there is no direction in space along which all the electrons have the same spin-projection. (For finding whether the electrons are polarized, and on ...

2

The density matrix formalism follows naturally from a state vector formalism when we average over a part of variables. It is not a "more general", but a more common approach in practical applications. Sometimes it is the only possible formulation due to experimental restrictions.

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The Hilbert space of the states in this case is the Fock space. It is a linear space "constructed" by acting by the creation operators $a^\dagger_k$ on the vacuum state $|0\rangle$, which has the property $a_k|0\rangle=0$. All other states are related so that the commutation relations between $a,a^\dagger$ are satisfied. The individual states like ...

2

Usually in many body theory these operators create and annihilate particles. There are different annihilation and creation operators for fermions and bosons (they obey different commutation relations). The states they act upon and the states "created" by them respect the required symmetries (antisymmetric for fermions, symmetric for bosons). The operators ...

2

The notation $\lvert \phi \rangle \lvert \psi \rangle$ is shorthand for $\lvert \phi \rangle \otimes \lvert \psi \rangle$. What you are doing is flipping a tensor product around. Though, in general, $A \otimes B = B \otimes A$, it is not the case that $a \otimes b = b \otimes a$ for $a \in A, b \in B$ (since the left and right hand side don't even live in ...

1

When you change the free field $A_\mu$ by means of a gauge transformation, you can easily see that it affects longitudinal and timelike degrees of feedom. Since observables are gauge invariant, those degrees of freedom cannot be physical.

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The total field consists of the "near" field like the Coulomb one and more generally (and loosely) a retarded Coulomb field, which are always "attached" to the charge, and the photon (radiated) field with different polarization orientations. The near field is always present, its "photons" are not created and annihilated. The corresponding "photons", when ...

0

Yes, if $\hat{O}$ commutes with $\hat{H}_0$, $\hat{O}$ is diagonalized in the base of eigenvectors of $\hat{H}_0$. Let's show this. If the operator $\hat{O}$ commutes with $\hat{H}_0$, we have $$\hat{O} \hat{H}_0 = \hat{H}_0 \hat{O}. \qquad (1)$$ Let's assume for simplicity that the spectrum of $\hat{H}_0$ is non-degenerate. Now, let $V_1$ be an ...

0

Using $\hbar=1$, your matrix $$j_x=\frac{1}{\sqrt 2}\pmatrix{0 & 1 & 0\\ 1 & 0 & 1\\ 0 &1 &0}$$ has the igenvalues $1,0$ and $-1$ as you stated. But when you wnat to get the eigenvectors, you must use the factor $\dfrac{1}{\sqrt 2}$. That is, the system you have to solve is $$\pmatrix{-1 &1/\sqrt 2 & 0\\1/\sqrt 2 &-1 ... 3 The energy eigenstates can be expressed in the form of wavefunctions as well, e.g. \psi_n(x) \equiv \langle x | E_n \rangle. Then, you can compute the inner product of the two wavefunctions by integrating their product:$$\langle E_n | \psi \rangle = \int_{-\infty}^\infty \langle E_n|x\rangle \langle x|\psi \rangle \, dx = \int_{-\infty}^\infty ...

1

$$\sum _nf(n)\int \mathrm{d}p\, |\left< p|n\right> |^2=\sum _nf(n)\int \mathrm{d}p\int \mathrm{d}q\, \left< n|p\right> \left< p|n\right> =2\pi \sum _nf(n)\left< n|n\right> =2\pi \cdot \sum _nf(n),$$ where I used the fact that $$\int \mathrm{d}p\, \left| p\right> \left< p\right| =2\pi,$$ which follows from $$\left< ... 1 Your claim that the derivative is in the expansion of \langle x'|\hat{p}\hat{x}|\psi\rangle acts on everythin on the right is correct. Realize that$$\langle x|\hat{p}|\psi\rangle=\frac{\hbar}{i}\frac{d}{dx}\langle x|\psi\rangle=\frac{\hbar}{i}\frac{d}{dx}\psi(x)$$So$$\langle x'|\hat{p}\hat{x}|\psi\rangle=\frac{\hbar}{i}\frac{d}{dx'}\langle ...

1

$a^\dagger a\! \mid \! n \rangle = n\mid \! n \rangle$. For vacuum you can write $a^\dagger a\! \mid \! 0 \rangle = 0\mid \! 0 \rangle$, if you like. No need to write 0 solely.

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It means it's "the end of the line". The vacuum state is, as you correctly say, not the zero state. It has energy content, and physical meaning - it's the state with no particles. Annihilating the vacuum leaves...nothing. Trying to take a particle out of it is not possible - it gives you the zero vector, which does not represent a physical state, since it is ...

0

this is just a change of basis. Its like a finding a new set of base vectors after rotating a coordinate system in 3 space. For your example you have infinite base vectors not three. For the negative of in front of the ket its just like multiplying a base vector by some constant. -1 in this case. it would make more sense to write -1 times the ket then ...

9

In a nutshell: With a couple caveats: what is plotted here is really the wavefunction $-\langle x\lvert 2\rangle$ or $\langle x\lvert -2\rangle$ (so actually this graph is highly misleading in one way), and the spikes should be thought of as delta functions. But this is the basic idea.

4

When dealing with the bra-ket notation, we must be clear that what we put inside the bras or kets is just a label and not something we make math with, although good labels makes visualization easier. When you write $|x\rangle$ you are considering the state of the particle in the $x$ position. (There is a problem here that this is not an acceptable state, ...

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$\def\ket#1{\left|#1\right>}$Each of those kets represents the state of a system. In quantum mechanics the constant multiplying a wavefunction is not physically significant (multiplying a wavefunction by a number does not change the physical meaning of the wavefunction). So if $\ket{x}$ represents a wavefunction completely localized to the point $x$, then ...

2

As correctly pointed out by Daniel Sank in the comment section, the key to understanding the state space in quantum field theory is the realization that it contains information about the excitations of operator-valued functions (quantum fields) of spacetime. The latter consists of one time and three spatial coordinates (at least in the context of the ...

-1

States in QFT are not so relevant, because what you are calling a "state" doesn't contain the physical information of the system in QFT. The relevant quantities in QFT are the correlation functions, which allow us to calculate explicitly transition probabilites between configurations of the same or different number of particles (the latter is the main goal ...

0

To put it simply, a QFT state is a linear superposition of the possibilities of one-particle wavefunction, two-particle wave function, three-particle wavefunction, ad infinitum. Each of those wavefunctions is a map from $\mathbb{R}^3 \rightarrow \mathbb{C}$ just like what you said. EDIT: Check @Alvaro's updated answer. BTW, the time evolution of that ...

0

If I know the explicit form of the states I would first apply the SU(2) generators to them. For example apply the Cartan generator $L_3$ and see whether you get the eigenvalues $-5/2, -3/2,...,+5/2$.

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