# Tag Info

1

You do not obtain the rules for the infinite-dimensional case by "proving" them from the finite-dimensional rules. Rather, you know that you need to have a Hilbert space, which is a complex vector space with an inner product, essentially. If you now search for infinite-dimensional Hilbert spaces that could possibly be used in quantum mechanics, you find ...

0

The correct formulation of the result is called spectral theorem. It encompasses all type of spectra for self-adjoint operators, and thus both (1) and (2). Given a self adjoint operator $A$ with only discrete spectrum (you have such situations for compact operators, and operators with compact resolvent) the spectral theorem has a nice and simple form: ...

0

For describing the behavior of a particle we usually write the Schrodinger equation, or Dirac equation, or other. All the possible solutions of the equation form a space of functions with certain properties, and we name such a space of functions Hilbert space. Now, if we have two particles, $1$ and $2$, which don't interact, behave as if none of them is ...

2

Let $$\tag{1} \hat{T}_{ik}~:=~\hat{n}_i \hat{n}_k-\frac{1}{3}\delta_{ik}\hat{\bf 1}.$$ The phrasing of the problem in Ref. 1 is indeed not the clearest, but by comparing with the given solution, it seems that Ref. 1 is performing a partial averaging over the Hilbert space of states with fixed value of the orbital angular momentum quantum number $\ell$ and ...

0

Apologies, I did not read the question very clearly. I leave my old answer below as it answers directly the title and therefore may help future comers. This does appear to be slightly sloppy language (though I wouldn't necessarily blame it on L&L if it's not present in the Russian original, but my Spanish copy has an equivalent form).I would read the ...

1

The trace class operators form a Banach space. There is a concept of (countable) basis for Banach spaces that is called Schauder basis. Not every Banach space has a Schauder basis, but it is true e.g. for the space of compact operators (the case of $\mathcal{K}(l^2)$ is given explicitly in the wikipedia article). Since the trace class operators are ...

0

This may not be exactly what you want, but it's something. The definition $$\vec{L}=\vec{r}\times\vec{p}$$ combined with $\vec{p}=-i\hbar\vec{\nabla}$ implies that the the $z$ projection may be written as $$L_{z} = -i\hbar\frac{\partial}{\partial \phi}.$$ Where $\phi$ is the azimuthal angle. Since $L_{z}|\ell m\rangle = m\hbar |\ell m\rangle$, the $\phi$ ...

0

Every density matrix on $\mathbb C^2$ is of that form because $\rho$ must be Hermitian. This property is clearly satisfied because of the form and the algebra of Pauli matrices. This can then be generalised to higher dimensional Hilbert spaces by introducing a basis for Hermitian matrices which satisfy the same algebra. So the problem is then equivalent to ...

0

In case $\mathcal{H}=L^{2}(\mathbb{R}^{n},d\mathbf{x})$ $\rho$ is a positive semi-definite trace class operator and can be expressed as $$\rho =\sum_{k}\lambda _{k}|u_{k}><u_{k}|$$ where $\{u_{k}\}$ is a basis for $\mathcal{H}$ and $\lambda _{k}$ is non-negative with $$\sum_{k}\lambda _{k}=1.$$ Addition in response to a question by Norbert ...

5

If $A$ is self-adjoint, you can define $f(A)$ as a complex-valued observable, where $f: \mathbb R \to \mathbb C$ is a measurable complex-valued function: $$f(A) := \int_{\sigma(A)} f(x) dP^{(A)}(x)\:,$$ $P^{(A)}$ being the spectral measure (projector-valued) of $A$. $N= f(A)$ is a closed normal operator and admits a spectral decomposition $P^{(N)}$ ...

1

If the operator is not self-adjoint then this is a possibility. If you search on phys.SE you will find questions about the expectation value of xp in the case of the QHO, and this turns out to be imaginary

2

Comments to the question (v1): We will not discuss tachyonic states here, because they are pathological and signal an instability of the theory. Then $$\tag{1} p^{\pm}~\equiv~\frac{p^0 \pm p^1}{\sqrt{2}}~\geq~0$$ is manifestly non-negative, since the energy $p^0\geq |p^1|$. In the light-cone formalism $p^{+}>0$ is strictly positive, since the special ...

1

I am not giving a mathematical solution to your problem, as that can easily be found by a quick google search. Instead, I prompt you to imagine the following, in hopes that it furthers your intuitive understanding of the problem: Picture in your head a three-dimensional space. Take any one plane, e.g. the $x$-$y$-plane. Let, as an example, $P$ be the ...

0

A normal operator with spectrum given by just the point $0$ is the zero operator. Clearly there are projections which are not the zero operator, so you might want to revise your argument.

0

Actually, the conditions $$\langle x|A x \rangle \geq a \langle x |x \rangle$$ and $$\langle x|A x \rangle \leq b \langle x |x \rangle$$ with $a,b \in \mathbb R$ fixed and all $x\in D(A)$ (the domain of $A$) refer to boundedness (resp. from below or from above) of the quadratic form associated to the linear operator $A$. This operator can always be supposed ...

0

Mutually commutative means that every operator in the set commutes with every other one. This implies that, if the operators in question are observables, they can all be measured simultaneously. A complete set of mutually commuting observables is a set of observable, hermitian operators that commute - therefore their eigenvalues can be used to label a ...

1

No, the Hilbert space is not spanned by continuous "eigenfunctions" because they are not eigenfunctions at all! A self-adjoint operator $T$ on $L^2(\mathbb{R})$ has a point spectrum, a continuous spectrum and a singular spectrum. The latter is physically irrelevant. The point spectrum consists of the values $\lambda_i$ for which a true eigenvector ...

3

The expansion formally works for any operator. It's breaking the exponential as $$\exp(-x)=\frac{\exp(-x/2)}{\exp(x/2)}$$ and then expanding the numerator and denominator as $e^x\approx1+x$. However, since the exponential term in the Cayley expansion comes from the time-evolution of a wave-function: $$\psi(x,t+\delta t)=e^{-iH\delta t}\psi(x,t)$$ which ...

1

The identity you used, $$\exp(i\theta \, \hat s)=\cos(\theta)+i\sin(\theta)\,\hat s, \tag{\ast}$$ is crucially dependent on the operator $\hat s$ being idempotent, and particularly on the fact that $\hat{s}^2=\mathbb1$. This is generally not the case for angular momenta other than spin-1/2. In general, the total angular momentum is a scalar within the ...

1

The point is that the spin operator is defined to be (1/2) times SU(2) generator while the orbital angular momentum is defined to be only SU(2)(or SO(3), is the same) generator. The proof is the same, and is " representation independent", in the sense that the structure of identity multiplied by something plus a linear combination of sigma matrices ...

0

You can use similar derivation, but with momentum operator J for integer spin instead of Pauli matrix $1/2\sigma$. Then you get integer coefficient in from of $\alpha$ instead of half-integer in the final formula... and the same state after $2\pi$ rotation.

1

So, the problem is that you've got to enforce Fermionic antisymmetry, but Fock space tries to make things easier by making that invisible. So if we've got two electrons in a box in a definite Fock state, the electrons definitively occupy some single-particle states which we can just call $1, 2$. The actual state that is being occupied is therefore: ...

1

The anticommutation rules for creation/annihilation fermionic operators are what defines these operators. The "proof" that they are correct is that they produce a theory that is compatible with the antisymmetric nature of fermions (and with all the other experimental results of course). For example you can check that they produce the expected result for ...

0

The moller operator take the wave function and send to $t=-\infty$ with the influence of potential $V$ and send back to $t=0$ without the influence of $V$. $$\Omega_+ = \lim_{t \rightarrow -\infty } e^{i H t } e^{- i H_0 t }$$ If we make this with a far away wave function $\psi$, this wave function don't feel the potential $V$ at time $t=0$. But we don't ...

1

To say that something is a (linear) operator you have to specify the space where it acts. You may say that, for example, wavefunctions of quantum mechanics are maps: $t\to \psi(t)$ that are continuous in $t$ with values in $L^2(\mathbb{R}^d)$. If we restrict to compact time intervals $[0,T]$, we may denote the space of these maps by ...

2

In fact $xp$ is not self-adjoint, it can have non-real expectation values. But its symmetrized form $D=(1/2)(xp+px)$ is better behaved (it has a self-adjoint extension). It is the generator of dilatations which scales momenta and coordinates. The complexification of ${\exp}[iDa]$ (i.e. $a$ becomes complex) is important for the study of the spectrum of a ...

1

You are touching on the subject of relativistic quantum mechanics where time and space $(t,x)$ are handled on the same footing as operators. The accepted description is to not use quantum wavefunctions as describing one particle but rather the state of a quantum field. Doing this turns into the subject of quantum field theory and is the basis of modern ...

1

in short the first way, but you can see this via using the position basis, $$<x|\psi> = \psi(x) , \ 1 = \int |x><x| \ dx, \ p|x> = |x> (-i \hbar \partial_x), \ <x|x'> =\delta(x-x') \ \rightarrow \\ <\psi|xp|\psi> = \int dx \ dx' <\psi|x><x| x p | x'><x'|\psi> = \int dx \ dx' \psi^*(x) x <x|p | x'> ... 2 TL;DR: The property bounded, bounded from above, and bounded from below are different things, cf. Wikipedia. In detail, consider a densely defined symmetric linear operator A:D\subseteq H \to H in a complex Hilbert space H. Let$$\langle A \rangle_{\psi}~:=~ \frac{\langle \psi, A\psi\rangle}{||\psi||^2}$$for \psi\in D\backslash\{0\}. It follows ... 1 When one says that an operator is bounded, you can think of it as being bounded from above. This is different from being bounded from below. An operator can be bounded (from above) and bounded from below, or perhaps just bounded, or just bounded from below. Observe that (Af,Af) and (\psi,B\psi) are slightly different: the former is always positive for ... 1 Let D be the subspace of Schwartz functions \psi of rapid decrease \mathscr{S} such that their primitive \Psi is in \mathscr{S}. Then A:D\to D (easy to see calculating the primitve with integration by parts); A defined on \mathscr{S} and B:D\to \mathscr{S}. Hence on D both AB and BA are well-defined, and$$[A,B]\psi=B\psi\;,\; ...

0

The first term works out by the fundamental theorem of calculus. Edit: Deleted my point about square integrability.

1

The time evolution of the two spins can be separated if they are independent, i.e. if they don't interact. Under this assumption the time operator splits in the tensor product $$U_1\otimes U_2=(U_1\otimes I)(I\otimes U_2)$$ and therefore it is clear how to define the time evolution for the single spin: for the $j$th particle one simply needs to take the ...

1

An observable that has a definite value is an eigenvalue of the operator. If $A$ is a hermitian matrix of which $|\psi\rangle$ is an eigenstate, we have $$\tag{1}A|\psi,a\rangle=a|\psi,a\rangle$$ You asked about the wave function, not the state vector, though. We can still get all the information we need from the wave function \$\psi(x)=\langle ...

0

How to prove that the position operator in momentum is iℏ∂/∂p Apply the following useful result. If $$F(k) = \int_{-\infty}^{\infty} f(x)e^{-ikx}dx$$ then $$\int_{-\infty}^{\infty} xf(x)e^{-ikx}dx = \int_{-\infty}^{\infty} f(x)\left(i\frac{\partial}{\partial k} e^{-ikx}\right)dx = i\frac{\partial}{\partial k}\int_{-\infty}^{\infty} f(x)e^{-ikx}dx = ... 2 (\hbar omitted in the following.) That is not weird, it is one of the crucial properties of the Fourier transform F(\bar{}) that$$ F(\partial_x f) = \mathrm{i}p F(f)$$i.e. differentiation by one variable becomes multiplication with the Fourier conjugate variable and vice versa. Because of this, Fourier transformation is a powerful tool to solve ... 0 Coordinates and momenta of the system are called variables by convention. Time is sometimes called variable, after all its nature is to keep changing. But to distinguish time from coordinates and momenta, people call it parameter in some cases. Parameter is a word used instead of variable when the quantity is a different kind of argument of a function. ... 3 The formula that Ref. 1 uses is$$\tag{*} \exp\left(-\sum_j \eta_j a_j^{\dagger} \right) ~=~ \prod_j\exp\left( - \eta_j a_j^{\dagger} \right) ~=~\prod_j \left(1- \eta_j a_j^{\dagger}\right). $$Ref. 1 correctly applies [the Hermitian conjugate of] eq. (*) to the bra in answer (a) on p. 181. There is no mistake on p. 181. Ref. 1 does not write a ... 0 For a univariable grassmann number \eta_1 it holds \eta_1^2 = 0 because of  \{ \eta_1, \eta_1 \} = 0. Hence all higher powers of this Grassmann number vanish. However, if there are multiple Grassmann numbers, linear combinations of these Grassmann numbers can be computed. It still holds \eta_i^2 = 0 for every i, but \eta_i \eta_j \neq 0 for i ... 1 If you want to write a super-operator representing left- or right-multiplication, there is a distinct method which is simpler and more elegant. Let us define the left-multiplication superoperator by$$ \mathcal{L}(A)[\rho] = A\rho,$$and the right-multiplication superoperator by$$ \mathcal{R}(A)[\rho] = \rho A.$$It should be clear that these operations ... 1 This is exactly analogous to the procedure for finding matrix elements of normal operators. Let's first recall how this works in the familiar case. You choose an orthonormal basis of vectors, say |n\rangle, with n = 1,2,\ldots D, where D is the dimension of the Hilbert space, such that \langle n\rvert m\rangle = \delta_{mn}. Now the matrix elements ... 1 The problem of including time as an operator rather than a parameter in Quantum Mechanics is what led to the development of Quantum Field Theory. I.e., the position operator was demoted to a parameter rather than promoting time to an operator. The two uncertainty principles you quote are entirely different. The first (position/momentum) principle is the ... 0 I think the question about "time" as an operator is quite controversial. It is not generally accepted that time can be associated to a (possibly unbounded) self-adjoint operator. One could naively argue that the time occurs in physical laws as a parameter rather than something we actually observer, although there clearly are instruments that can measure ... 2 A self-adjoint operator T on L^2(\mathbb{R}) has in its spectrum three different kinds of subspectra: A discrete point spectrum, a continuous spectrum, and a singular spectrum. The latter is physically discarded. The point spectrum consists of the eigenvalues of T, that is, the spectral values for which true eigenvectors in L^2(\mathbb{R}), and ... 2 Eigenvectors exist only for the point spectrum of an operator. For any other point of the spectrum one can only find a sequence of vectors for which (A-\lambda I)u_n\to0, where A is said operator, and \lambda is a point in the spectrum which is not an isolated point. So in this case there is a sequence of approximate eigenvectors. With a bit of extra ... 1 First of all, I would encourage you to think of position of acting on momentum states to the left, that is, to commute them with the bra:$$ ⟨\mathbf p|\hat x=-i\hbar \frac{\partial}{\partial p_x}⟨\mathbf p|, $$where the differentiation is over anything to the right of it, so for instance$$⟨\mathbf p|\hat x|\psi⟩=-i\hbar \frac{\partial}{\partial ...

0

wich I think is nonsense Formally, shouldn't the 2nd equation be $$\langle x \rangle = \frac{\langle\psi_{\mathbf p_0}|\hat x|\psi_{\mathbf p_0}\rangle}{\langle\psi_{\mathbf p_0}|\psi_{\mathbf p_0}\rangle} = \frac{f^2(p_0)\int d\mathbf x^3 x}{f^2(p_0)\int d\mathbf x^3} = \frac{\int d\mathbf x^3 x}{\int d\mathbf x^3} = \lim_{\tau \rightarrow ... 1 Nothing will help, a plane wave occupies all the space, and the mean value of the position doesn't make much sense, but in your case is zero, because the integrand in the 1st calculus is anti-symmetrical. But this, on one condition, namely if we write the integral as lim _{a \to \infty} f^2(p_0) \int _a^a x d^3 x Otherwise it's hard to say what is the ... 0 Using the comment of @MarkMitchison, since e^\hat{C}=\sum_{k=0}^\infty\frac{\hat{C}^k}{k!} and e^x=\sum_{k=0}^\infty\frac{x^k}{k!} (as can be seen here), so e^{\alpha\hat{I}}=e^\alpha\hat{I}, and I "can take the first term outside the trace in Eq. (2). You don't need to assume any special properties of the trace". Again, thanks @MarkMitchison. 16 I understand the definition of a Hilbert space. But I do not understand why non-commutativity compels us to use Hilbert spaces. It doesn't, but that's not what Scrinzi is saying. The reason is doesn't is because we could work, for example, in Wigner quasiprobability representation:$$\rho\mapsto W(x,p) = \frac{1}{\pi\hbar}\int_{-\infty}^\infty\langle ...

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