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5

The field operator can be divided in two parts, one with positive frequency and other with negative frequency $$\phi(x) = \phi^+(x) + \phi^-(x)$$ $$\phi^+(x) = \int \frac{d^3p}{(2\pi)^3\sqrt{2\omega_p}} a_p e^{-ipx}\qquad \phi^-(x) = \int \frac{d^3p}{(2\pi)^3\sqrt{2\omega_p}} a_p^\dagger e^{ipx}$$ As you can see, the positive frequency part $\phi^+(x)$ is a ...


5

The inner product is always between two (ket) vectors. However, let $\mathcal H$ be the space in which they live. This is a complex vector space with a Hermitian inner product. The inner product defines a map $\mathcal H\to\mathcal H^\vee$, where $\mathcal H^\vee$ is the dual of $\mathcal H$, and consists of linear functionals on $\mathcal H$. The map is ...


4

Perhaps its a little clearer if you shorten the contents of the brackets (and lets drop the constants too): $$\frac{d\langle x\rangle }{dt} \propto \int _{-\infty} ^{\infty} x \frac{\partial }{\partial x} \left[ \ldots\right] dx$$ $$ = \int _{-\infty} ^{\infty} \frac{\partial }{\partial x} \left(x \left[ \ldots\right] \right)dx - \int _{-\infty} ^{\infty} ...


4

http://arxiv.org/abs/quant-ph/9607007 discusses necessary conditions on $T$ (more precisely, on its singular values) for $\rho$ to be positive. They don't seem to derive sufficient conditions, however. The basic idea is that one can perform a rotation $U_A$ and $U_B$ on the two qubits, respectively, which correspondingly transforms $r\mapsto O_Ar$, ...


4

By definition, a value is in the continuous spectrum of $A$ if it not an eigenvalue, but the range of $A-\lambda I$ is a proper dense subset of the Hilbert space. There is nothing in this definition distinguishing separable spaces, or precluding operators in them from having continuous spectrum, and indeed some do. There is an equivalent definition in terms ...


3

The visualization method you choose is directly and completely determined by the information you need to see regarding your state. For the states of a single bosonic mode, there are multiple different visualization methods, and they all have their pros and cons. In particular, there is a direct trade-off between the amount of information you can display on a ...


3

As noted, many people use "complete" where perhaps they ought to say "complete and orthogonal and orthonormal" or the like. I'm not sure what I can tell you besides confirming that usage is not always ideal. I'll answer one question you brought up, but I'm worried I may have gotten confused myself by what kind of "complete" you meant: Is it even possible ...


2

You need to be careful with the word span. A mathematician will say that the span of a set of vectors is the set of finite linear combinations, so you can only add linear combinations of finitely many at a time to get something in the span. So there are sets that are mutually orthogonal and all normalized but not enough to span the space with finite linear ...


2

The antiunitary operator is an operator on the Hilbert space. Thus, it is nonsense to say that "the operator $K$ apply on complex number". Nevertheless, it can be shown that $U\alpha |\phi\rangle=\alpha^*U |\phi\rangle$, where $U$ is an antiunitary operator. $U|\phi\rangle$ should not be confused with $\langle\phi|$, the complex conjugate of $|\phi\rangle$.


2

In the case of an infinite superposition of eigenstates it becomes more complicated but we can still write a general expression for it. If $$\psi(x) = \sum_{n=0}^\infty a_n \phi_n(x)$$ where the $\phi_n$ are the eigenstates of the Hamiltonian. The time dependent wavefunction will look like: $$\Psi(x,t) = \sum_{n=0}^\infty a_n \phi_n(x) T_n(t)$$ where $T_n = ...


2

In general, the integral $$ V := \int \mathrm{d} \mu = \int 1 \mathrm{d}\mu$$ is the integration of the identity over the space the measure $\mu$ is defined on, and should be intuitively understood as the volume of the space with respect to the measure. (This is usually only finite for compact spaces.) Normalizing the measure means sending $\mu \mapsto ...


2

The notion of separability of a state has a precise and simple meaning: In natural language, a separable state of a system that has several subsystems is a state to which a unique state of every subsystem is associated. In classical mechanics, all states are separable in this sense - given two configuration spaces $Q_1,Q_2$, the configuration space of the ...


2

Wick's theorem tells us that $$ \mathcal{T}(\phi_1\dots\phi_N) =\ :\phi_1\dots\phi_N: + :\text{pairwise contractions}:$$ where $:\ :$ is normal ordering. Immediately from the definition of normal ordering (all annihilators to the right, all creators to the left), the expectation value of anything that is normal-ordered and not a constant vanishes because the ...


2

$\newcommand{\ket}[1]{\left| #1 \right>}$If your state is in an eigenstate of the energy operator then the answer is that you'll get the same value for the energy every time you measure the particle's energy. That is the reason why the energy eigenstates are also called stationary states. On the other hand you can also have a superposition of energy ...


1

Are you sure that's what the book is asking you to find? $\hbar/2$ is the eigenvalue of the $S_{x}$ operator corresponding to spin up, but it is not part of the state vector. If the question is really asking you to express the $\mid S_{x};+\rangle$ ket in the $S_{z}$ basis, then you're nearly correct, just a minor sign error: $$\mid S_{x};+\rangle = ...


1

Integration over pure quantum states usually refers to the Haar measure, i.e., the unitarily invariant measure. Vaguely speaking, you assign the same volume (=weight in the integral) to any two set of states which are related by an arbitrary unitary rotation $U$. In the case of one qubit, this is equivalent to integrating over the Bloch sphere; i.e., we ...


1

I think you could work it like this: $X_+ ={1 \over \sqrt{2}} (\begin{matrix} 1 \\ 1 \end{matrix}) =a (\begin{matrix} 1 \\ 0 \end{matrix} ) +b(\begin{matrix} 0 \\ 1 \end{matrix} ) $.where $X_+$ is the eigenvector on the positive axon of $S_x$ Solve and find a,b and there you are. Note also that you can write a general spinor as $(\begin{matrix} cos\theta ...


1

If we consider local operators $M_A$ and $N_B$ acting on Alice's and Bob's part, respectively, then it holds that $[M_A,N_B]=0$, i.e., we can use this property in proofs involving local operations. Note that conversely, however, commutativity need not imply locality (to start with, there need not even be a tensor product structure).


1

The word separable (the property) has a precise and detailed meaning (about lack of factorizability) when discussing multiparticle states. But that meaning might not be the meaning in the context you consider. For instance the paper http://arxiv.org/abs/1302.7188 argues that separability (the principle) is not related to Bell's inequality. And in that paper ...


1

No, that is not quite correct. The Schrödinger equivalence says that if the system has a definite energy, then this energy can only be an eigenvalue of the system's hamiltonian $\hat H$. There is no requirement, however, for the system to have a definite energy; if the energy is undefined then an energy measurement may return different (eigen)values on ...


1

I am not at all an expert of quantum gravity, but I think you have misunderstood the point. As I understand it, the point is not necessarily having distributions as quantum Hilbert space vectors, but having a distributional "configuration space", i.e. distributions as the domain of the function(al)s that are the quantum vectors. While in QM (i.e. for ...


1

The derivative of $x f(x)$ is $x f'(x) + f(x)$, and here you're integrating $k \int_{-\infty}^\infty dx ~ x ~ f'(x)$ for some constant $k$, and some complicated function $f$. When you integrate this by parts, you raise $f' dx$ and lower $x$ to find:$$k \int_{-\infty}^\infty dx ~ x ~ f'(x) = k \left[x ~f(x)\right]_{-\infty}^{~\infty} - k \int_{-\infty}^\infty ...


1

The question is really one of definition. In the math literature on self adjoint opertors the "discrete spectrum" is by definition that part of the spectrum which consists of normalizable states, while the "continuous spectrum" is that part where they are non-normalizable. It is possible to have a physical system (a random potential on the entrire real ...



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