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2

An intuitive answer to motivate the Stone theorem that ACuriousMind's answer cites. Take a quantum system. For now, let it be a finite dimensional one (the Stone theorem is needed to make the reasoning work in a separable, infinite dimensional Hilbert space). Let it evolve for time $t_1$. The evolved state is some linear operator $U(t_1)$ imparted on the ...


3

A very simple proof proceeds along the following lines. $$\Psi(t_1+\Delta t)=\Psi(t_1)+\frac{\partial \Psi}{\partial t} \Delta t $$ where we omitted higher order $\Delta t$ terms (will disappear when we let $\Delta t \to 0$ later). Applying this formula recursively, we get $$\Psi(t_1+2\Delta t)=\Psi(t_1)+2\frac{\partial \Psi (t_1)}{\partial t} \Delta ...


5

This rather detailed answer will potentially chart out the issues and assumptions that are needed for a straightforward idea. Understanding how it is done might be more useful than actually doing it. Assumption 0) The Hamiltonian is time independent. To be honest, I never find arguments about taking derivatives of operators very appealing. We will have a ...


0

The relation between $U=U(t_1,t_1+t)$ and $H$ is $$H={\partial U\over\partial t}\Big|_{t=0}$$


8

For time-independent Hamiltonians, $U(t) = \mathrm{e}^{\mathrm{i}Ht}$ follows from Stone's theorem on one-parameter unitary groups, since the Schrödinger equation $$ H\psi = \mathrm{i}\partial_t \psi$$ is just the statement that $H$ is the infinitesimal generator of a one-parameter group parameterized by the time $t$. For time-dependent Hamiltonians $H(t) = ...


1

If the eigenvectors of $\hat H$ don't depend explicitly on $t$ (the eigenvalues can) then we can easily make sense of $\hat H~U = i \partial_t U$ by working in the eigenbasis of $\hat H$ where $H$ is diagonal, because if $\hat H = \operatorname{diag}(\lambda_1, \lambda_2, \dots)$ then $\exp(-i~\hat H~t) = \operatorname{diag}(e^{-i\lambda_1t}, ...


1

This is actually a general principle not specific to phonons. Given a collection of creation operators $a^\dagger(k)$ creating 1-particle momentum eigenstates of momentum $k$, these states are not actually viable states - they do not lie inside the Hilbert space, since their "inner product" with each other is not a finite number, but usually a delta ...


0

Any one of operators Lx, Ly, or Lz can be called quantized. There exists a set of states which are eigenstates of Lz; the matrix Lz is diagonal but Ly and Lx are not. There exists a set of states which are eigenstates of Ly; the matrix Ly is diagonal but Lx and Lz are not. There exists a set of states which are eigenstates of Lx; the matrix Lx is diagonal ...


0

If $[A,B]=0$ then there is a unitary transformation $U_{A,B}$ that diagonalises both $A$ and $B$ simultaneously. This transformation depends on the pair of commuting operators $(A,B)$, so that for a different pair there could be a different unitary. Assume that all the eigenvalues of $H$ have multiplicity 1. Then there exists a unique (up to a phase factor) ...


1

If $H$ commutes with $A_1$, then it will indeed share an eigenbasis with it. Your mistake is in supposing that it will share the same eigenbasis with both $A_i$s. Examples are easy to provide: On the trivial side, if $H=E_0\mathbf 1$ is trivial, then it shares an eigenbasis with $A_1=x$ and it shares an eigenbasis $A_2=p$, but it cannot share an ...


1

A 1-particle Hilbert space (neglecting spin for simplicity) is usually modelled as $L^2(\mathbb R^3)$ which is a function space, and the Hamiltonian is a differential operator. N-particle Hilbert spaces are usually constructed as tensor products of this 1-particle Hilbert space, but there exists an isomorphism such that you can again interpret them as ...


4

The expression $$ \hat{H}=\sum_j \varepsilon_j\,a_j^\dagger a_j $$ is not the most general expression for free particles hamiltonian because it implies that you already found the eigenvalues $\varepsilon_j$ and diagonalized $\hat{H}$, i.e. already solved the Schrödinger equation. Maybe you should look at the problem in a different basis. Let say $\{\vert ...


0

The second-quantization hamiltonian you wrote above is the hamiltonian for a collection of independent harmonic oscillators, where the quantities oscillating are now the amplitudes of the normal modes of the field being quantized. So, yes, the hamiltonian is a differential operator on the space complex functions of $N$ variables, one for each normal mode. ...


0

For bosons: We put ourselves in a suitable common domain, i.e. the finite particle vectors. Then $$\bigl(a^*(f)\Psi\bigr)_n(X_n)=\frac{1}{\sqrt{n}}\sum_{j=1}^n f(x_j)\Psi_{n-1}(X_n\setminus{x_j})\\ \bigl(a(f)\Psi\bigr)_n(X_n)=\sqrt{n+1}\int \bar{f}(x)\Psi_{n+1}(x,X_n)dx$$ Hence by definition $$\bigl(a(g)a^*(f)\Psi\bigr)_n(X_n)=\sum_{j=1}^{n+1}\int ...


5

Applying $\partial_1\overline{\partial}_1$, we have the following. First term:$$\partial_1\overline\partial_1 G_1 = -\pi\alpha' \eta^{\mu_1\mu_2} \delta^2 (z_1-z_2,\overline z_1-\overline z_2) X^{\mu_3} (z_3,\overline z_3) X^{\mu_4} (z_4,\overline z_4) $$$$+\text{ } {\rm permutations~of~indices~} (2,3,4).$$In the second term, $z_1$ can be the logarithm or ...


1

If we're dealing with a finite-dimensional Hilbert space, then the answer appears to be that you can always find a way to label the eigenvalues such that they are all differentiable (at least). See this paper and references therein: A. Parusinski & A. Rainer, "A New Proof of Bronshtein's Theorem". In particular, Theorem 2.4 of the paper states ...


3

The proof is given in great generality by the beautiful and powerful spectral theorem; in its functional calculus form. This theorem clarifies the meaning of your first expression, and specifies the functions for which you can write the second (that however are many). In addition, it applies to any self-adjoint operator, bounded or unbounded. Sadly, the ...


3

First of all you need the vectors $|O_k\rangle$ to form an orthonormal system. Then to prove what you need you start with polynomials and verify that, e.g. $O^2$ has the desired form etc... For a continuous function $f$, assuming that $O$ is a bounded operator you can then invoke the Stone-Weierstrass theorem.


5

A first remark: the term "Hermitian", even if very popular in physics is in my opinion quite misleading (because someone uses it for symmetric operators, others for self-adjoint ones). A second remark: the self-adjoint operators of a given Hilbert space $\mathscr{H}$ are in one-to-one correspondence with the strongly continuous groups of unitary operators; ...


1

Most of the time, an (elastic) scattering problem can be reduced in : An incoming initial wave / quantum state $|\phi\rangle$, which most of the time is taken to be a plane wave / free state $|\textbf{k}\rangle$ eigenstate of the free hamiltonian : $$ ...


2

For normal elements in a C*-algebra you can do continuous functional calculus, that is, if $a$ is a normal operator, then $f(a)$ is well-defined for any $f\in C(\sigma(a))$. Since $\sigma(a)$ is always compact you can use Stone-Weierstrass to write $f$ as a uniform limit of polynomials in one complex variable and its complex conjugate. Hence you can verify ...


2

Perhaps the easiest way to see that there should be a Grassmann sign factor $(-1)^{|A| |B|}$ in the definition of time ordering $$\tag{1} {\cal T} \left\{ A(t_A) B(t_B)\right\} ~:=~ \theta(t_A-t_B) A(t_A) B(t_B) + (-1)^{|A| |B|} \theta(t_B-t_A) B(t_B) A(t_A), \qquad $$ is to go to the classical limit $\hbar\to 0$. Here $|A|$ denotes the Grassmann parity, ...


3

Let's calculate $[L_x L_y, L_z]$. I'm going to use the property $[AB,C]=A[B,C]+[A,C]B$. If you apply the property to our case, you obtain $L_x[L_y,L_z]+[L_x,L_z]L_y$. Now you can substitute the value of the commutators and find the correct answer. Note that the quantum commutation relations are pretty similar to vector product in cartesian coordinates. For ...


1

The NOT gate can be written as $$ \sigma_x \propto \exp[i\pi/2\,\sigma_x] $$ (up to an irrelevant global phase). Its roots are therefore of the form $$ \exp[i\phi\,\sigma_x] = \cos(\phi)\,I + i\sin(\phi)\,\sigma_x $$ with $n\phi = \pi/2 + 2\pi k$ with $k\in\mathbb N$.


1

This topic seems to me as functions of operators, so I'll explain this issue and will use "n-th root NOT gate" as example. Any function you apply on the operator - is applied on it's eigenvalues. If the operator is diagonal - all the eigenvalues are on the diagonal and applying the function is simply apply it to any element on the diagonal. $$A=D=\Sigma_i ...


1

I'm guessing that by $n^{\text{th}}$ root you mean some gate which if you pass a signal through $n$ of them the output will be identical to the output of the gate you want to "root". For odd $n$ it's pretty obvious that the classical NOT gate is it's own $n^{\text{th}}$ root. In that an odd number of NOT gates one after the other are identical to a single ...


-3

commuting operators are any two operators which can be applied to a function in any order without altering the outcome


1

So the problem here is that you are confusing quantum mechanics of a single harmonic oscillator with quantum field theory, which is quantum mechanics of a field and can also be considered as quantum mechanics of an infinite number of harmonic oscillators. In quantum mechanics of a single oscillator, the ground state $|0\rangle$ can be represented as: ...


1

No, it has not discrete spectrum (on $L^2(\mathbb{R}^d)$). In fact $a+a^*$ is proportional to the position operator (or the momentum one, depends on your definition of $a$ and $a^*$; by the usual one the position operator $x$ is proportional to the real part $a+a^*$ and the momentum $p$ to the imaginary part $\frac{1}{i}(a-a^*)$). Both position and momentum ...


1

The answer to the problem is that the operator used to change to the new frame, $U(t) = D(\alpha)$ is dependent on time through the time dependence of $\alpha(t)$ and for the purposes of the derivation this time dependence is kept quite general. This means that we cannot use the usual rewriting of the equation of motion, but eq. 81 in the lecture notes ...



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