# Tag Info

5

For the absolutely continuous part of the spectrum of a self-adjoint operator $H$, the "density of states" is provided by the Radon-Nikódym derivative of the spectral measure of $HP_{ac}$ with respect to Lebesgue measure, where $P_{ac}$ is the orthogonal projection onto the absolutely continuous subspace of the domain of $H$. This formula is well defined ...

1

Momentum and position are conjugate variables in classical mechanucs, which means they satisfy the Poisson bracket relationship. When quantum mechanics was invented the Poison bracket relation was replaced by the operator commutation relationship which results in the relation under consideration.

3

I am not sure I understand perfectly you question but formally in the canonical ensemble we can write the partition function $Q(\beta)$ as being: $$Q(\beta) = \int\cdot \cdot \int d\mu(x)\: e^{-\beta H(x)} = \int_0^{+\infty} dE \: \rho(E)e^{-\beta E}$$ where $d\mu(x)$ is the volume measure for the micro states in the system, ...

-1

Ab initio the momentum operators can be constructed using de Broglie Plane waves In one dimension, using the plane wave solution of the Schrodinger equation,the wave function Psi = exp. i (kx -wt) , if one takes the partial derivative w.r. to x of the wave function delta/delta x (Psi) = ik. Psi and using de-Broglie relation p = hbar . ...

-1

Momentum is the generator of spacial translations, even in classical physics. Anyway, you can find a derivation here or in Sakurai's book Modern Quantum Mechanics. They are more or less the same and go like this: The translation operator is the operator $T( a)$ such that $$T( a) \mid x \rangle = \mid x+a\rangle$$ From the definition it follows that the ...

0

Theoretical physics is about constructing a mathematical model which we hope describes the phenomena it's being modeled for and hence helps predicting stuff. In classical physics this mathematical model is based simply on the real numbers (at least locally) because of the nice behaviour of things. In quantum mechanics it's not the case. Experiments started ...

0

That $\hat{P} = - i \hbar \partial_x$ generates translations comes from a straight-forward computation: if $\psi$ is continuously differentiable, and $\Psi$ as well as its derivative are square integrable, then you can prove that \begin{align} i \frac{\mathrm{d}}{\mathrm{d} y} \bigl ( \psi(x - y) \bigr ) \big \vert_{y = 0} = - i \partial_x \psi(x) ...

4

Historically, you probably want to start with the de Broglie relations (i.e. $p = \hbar k$), which are just a wild guess. This immediately pops out the form of $p$ as an operator if the wavefunction is a plane wave. Mathematically, $p$ should be defined as the generator of translations (or equivalently the conserved quantity corresponding to translational ...

0

This is standard material for any text on 2d CFT's. First, the Virasoro algebra contains holomorphic and anti-holomorphic parts and you can see (exercise!) that the conformal blocks factorize into respective pieces. So lets consider just the holomorphic part. The key is the decomposition of the identity in the conformal module of the intermediate state (the ...

0

No, it's not true and a simple counter example suffices. Let $A = \vec L^2$, the square of the angular momentum operator, and $B = L_z$, the z component of the angular momentum operator. $[\vec L ^ 2, L_z] = 0$, but $[\vec L , L_z] != 0$ because the components of the angular momentum operator do not commute with each other.

3

It's not true. For example take $$A = \begin{pmatrix} 0 & 1\\ 0 & 0 \end{pmatrix} \qquad\qquad\qquad B = \begin{pmatrix} 0 & 0\\ 1 & 0 \end{pmatrix}$$ you have $A^2=0$ so that $[A^2,B]=0$, but $$[A,B] = \begin{pmatrix} 1 & 0\\ 0 & -1 \end{pmatrix}.$$ You must add some condition, for example if you know also $$[A,[A,B]]=0$$ ...

2

If you ask about what plays the role of the state of the system at some given time than the answer is: nothing. You talk only about the initial state and what you get in the measurements (expectation values or probabilities of outcomes for observables). The Heisenberg picture is very "Copenhagen" in its spirit and abstracts itself from what's happening with ...

1

The physical states in the Heisenberg picture are frozen in time, and can be made to coincide with the Schrodinger-picture state at any given time $t_0$. In other words, $$|\psi(t_0)\rangle^S=|\psi\rangle^H$$ which doesn't evolve with time.

3

Let us assume that the operators are normalized so that $a^\dagger a$ has eigenvalues $0,1,2,3,\dots$. The prefactor in the exponent is $\mu h/2\pi=\mu\hbar$ must therefore be dimensionless, too. So $\mu$ has to have the units of the inverse action. Now, if $a=(x+ip)/ \sqrt 2$ – let's simplify the "masses etc." and set $\hbar=1$, then $$a^{\dagger 2} - a^2 ... 3 Similarly to AccidentalFourierTransform I am not sure to understand well your issue. However there is a crucial missed point in your argument, usually absent in many textbooks on these topics. It is true that decomposing H as H= \hbar\omega( a^\dagger a + \frac{1}{2}) and taking the relations (following from CCR) [a, a^\dagger] =I into account one ... 3 Well, I'm not sure I understood your question so I'm going to write what I think and let's see if it's useful :-) The algebra [a,a^\dagger]=1 is all you need to diagonalise H, but this is because what H looks like:$$ H=\omega a^\dagger a The important observables, namely H,P,X, can be written as polynomials in a,a^\dagger: \begin{aligned} ... 1 The adjoint of an operator is obtained by taking the complex conjugate of the operator followed by transposing it. i.e., (A)^\dagger_{ij}=\left((A)^T_{ij}\right)^*=\left((A_{ij})^*\right)^T=A_{ji}^* You can do it in any order. The adjoint of an operator is the infinite dimensional generalization of conjugate transpose, where you find the transpose of ... 2 What's special about H is that it generates time translations. This means that operators evolve, by postulate, through Heisenberg Equations of motion i\frac{\mathrm d}{\mathrm dt}\mathcal O(t)=[H,\mathcal O] $$modulo an explicit time dependence. Therefore, if an operator commutes with the Hamiltonian, [H,\mathcal O]=0 we automatically conclude that ... 2 Assume that g is a function in the space of rapidly decreasing functions with all their derivatives of any order, the so-called Schwartz space {\cal S}(\mathbb R^4). Define$$\hat{g}(k) = \frac{1}{(2\pi)^2} \int_{\mathbb R^4} e^{-i k_\mu x^\mu} g(x) d^4x\::$$As a consequence, as is well known, \hat{g} \in {\cal S}(\mathbb R^4) and$$g(x) = ...

2

PART 1 : Taylor Expansion ( see @Prahar comment ). In the following : $$\mathbf{x}=x^{\mu}, \quad \mathbf{k}=k_{\mu}, \quad \partial_{\mu}=\dfrac{\partial}{\partial x^{\mu}} , \quad \mu=0,1,2,3 \tag{1-01}$$ $$\mathbf{k}\cdot\mathbf{x}=k_{\mu}x^{\mu}=k^{\mu}x_{\mu}, \quad \text{Einstein's convention on}\: \mu \tag{1-02}$$ Suppose now that the ...

1

Here, $V_0(k_2)$ could have been replaced by $e^{k_2 \alpha_{-1}} e^{-k_2 \alpha_1}$ while the factors from $\alpha_{\pm n}$ for $n\gt 1$ could have been neglected because $\alpha_n$ annihilates everything that appears in the matrix element on the right side from $V_0$ (because it ultimately annihilates $|0\rangle$), and similarly for $\alpha_{-n}$ that ...

10

This is indeed possible only in some situations, e.g. when the continuous spectrum is absent (it may also consist of a single point, see Valter Moretti's comment below). A sufficient condition for that to be true is that either the Hamiltonian is compact or it has compact resolvent. Sadly, very few interesting Hamiltonians satisfy that property (an example ...

4

Heisenberg's uncertainty principle is in fact not a principle but a consequence of the operator formalism of QM. If we associate to the operator $X$ the standard deviation $$\Delta_X = \sqrt{ \langle{X^2}\rangle -\langle X \rangle^2}$$ it can be then shown that, given two operators $A,B$ $$\Delta_A \Delta_B \geq \frac{1}{2} \left| \langle ... 2 Yes, it is. The definition of the exponential of an operator is$$\exp(\hat X) = \sum_{k=0}^{\infty} \frac{\hat X^k}{k!} = \hat1+\hat X+\frac{\hat X^2}{2}+\dots$$where \hat 1 is the identity operator. So if you stop at the first order you will have indeed$$\exp(\hat A+\hat B) = \hat 1 + (\hat A + \hat B) $$-1 Yes it is as long as you define the exponential of an operator as e^X=\sum_{k=0}^\infty \frac{X^k}{k!} and A and B commute. If A and B do not commute it is correct to write e^Ae^B or e^Be^A instead of e^{A+B} since the order is important. In general$$e^Ae^B=e^{A+B+\frac{1}{2}[A,B]+\frac{1}{12}[A,[A,B]]+\frac{1}{12}[B,[B,A]]+\cdots},$$and ... 3 If |\pm\rangle are the eigenvectors of {\hat \sigma}_x, {\hat \sigma}_x |\pm\rangle = \pm |\pm\rangle, then a rotating x-basis is defined as$$ |+\rangle(t) = \exp\left(-i\frac{\omega t}{2} {\hat \sigma}_x\right)|+\rangle = e^{-i\omega t/2 } |+\rangle  |-\rangle(t) = \exp\left(-i\frac{\omega t}{2} {\hat \sigma}_x\right)|-\rangle = e^{i\omega t/2 ...

0

The author shows that since $UN$ equivalent to $UNU^\dagger U$, that it is possible to change the order of the operations, if $N$ is replaced by $M=UNU^\dagger$. In that case, $UN$ = $MU$. The remaining logic follows from this. Note that $M$ is just $N$ in the $U$ basis; the choice of $U$ uniquely determines how $N$ changes to $M$ for this ...

0

Concerning point 2: Operators do not always come through each other cleanly, but there are some very basic rules that always apply, which can be turned into less tedious rules that apply in special cases. Often the latter are taught first, causing mass confusion. General rule: Operators can be expressed as (Sum over a in the set of eigenvectors ) |a > ...

3

The annihilation operator is a linear operator. A linear operator can be applied to ANY state. And yes, it returns zero when applied to the ground state. You can really take this as a definition. The definition of the momentum operator $\hat{p}$ is the operator such that $\langle x|\hat{p}|\psi\rangle=-i\hbar \psi'(x)$. One could write this as $\langle ... 2 The point is that the domain$D(P)$of$P$must be such that$P$is (essentially) self-adjoint thereon. Otherwise it does not represent an observable. I am assuming that$D(X)= L^2([0,L],dx)$instead, where$X$is automatically self-adjoint. The vector$\psi$you use to prove Heisenberg inequality has to belong to$D(PX) \cap D(XP)$as you see by direct ... 2 The most effective way to think about such problems is by means of operator algebras. The best way to treat systematically quantum observables (and their associated unitary operators) is to collect them in a C$^*$-algebra, roughly speaking a Banach algebra where observables have a norm, can be summed and multiplied (with some additional technical ... 1 The answer is no, and the details are clearly spelled out in Glauber's Les Houches lectures (circa 1964). Glauber introduces a "T-representation" which can represent any operator in the Fock space of harmonic oscillator states, a less general "R-representation" which can represent any density operator, and the still less general "P-representation" which can ... -1 I wonder why we so seldom mention, when discussing these things, that you cannot answer this question without adopting a human convention with respect to the combination of spin states. If we take the +/- z direction to be the north and south poles, then any state with equal amplitude-squared in both components will correspond to a spinor pointing somewhere ... 1 First, let's clarify the expression of$|\phi\rangle$. The kets$|+\rangle_z$and$|-\rangle_z$are eigenvectors of$\hat{S}_z$such that $$\hat{S}_z |+\rangle_z = +\frac{\hbar}{2}|+\rangle_z \\ \hat{S}_z |-\rangle_z = -\frac{\hbar}{2}|-\rangle_z$$ This means that in the$\{|+\rangle_z = ...

0

If I'm understanding your question correctly, then the answer is that you actually can't justify the connection of your equations along the quantized direction. In fact, the "bands" in the direction perpendicular to your slab are going to be completely flat, which corresponds to an infinite effective mass in that direction. (The infinity comes from the ...

2

It's because always even number of Fermionic fields appear in the Hamiltonian. for example the Dirac Lagrangian for free electron: $\mathcal{L}=i\hbar\bar \psi(\gamma^\mu\partial_\mu-m)\psi$ has two $\psi$s. Invariance of the theory upon a global gauge transformation requires that each term in the Lagrangian have an even number of Fermionic fields. Since an ...

0

a Stern Garlach apparatus does not rotate the state of the particle, what it does is to split a beam or, if you have a single particle, it chooses" a state in the desired direction. What you might be looking for is how to write the eigenstate in terms of Z basis. For a 1/2 spin it is going to be: $$|\pm \rangle_y = \frac{1}{\sqrt{2}}\left( |+\rangle_z ... 2 Short answer to your question: yes. There's no reason for the path integral and the operator formalism to be equivalent. A simple example are all the non-Lagrangian theories. We know (some of) them through their operator algebra, but there's no corresponding Lagrangian. This might sound strange to most people, like myself not long ago, but it is not hard ... 1 Hints: The starting point is the 2-point relation$$T(\phi(x)\phi(y)) ~-~:\phi(x)\phi(y): ~=~ C(x,y)~{\bf 1}, \qquad C(x,y)~\equiv~\langle 0 | T(\phi(x)\phi(y))|0\rangle,\tag{1} $$cf. this Phys.SE post. The relevant Wick's theorem is a nested Wick's theorem$$ T(:\phi(x)^n::\phi(y)^m:)~=~\exp\left( ...

2

Everything you write is correct and there is no inconsistency. When you write $\left<x\right|Hf(x)g(x′)\left|x′\right>$ you just, indeed, put the numbers on the right. But numbers commute so there is nothign wrong with it. Note that you have really no conclusion from it. You can't even transform back to operators on the form ...

2

If $\mathbf{x}=\left(x_{1},x_{2},x_{3}\right)$ is a 3-vector rotated to $\mathbf{x}^{\prime}=\left(x_{1}^{\prime},x_{2}^{\prime},x_{3}^{\prime}\right)$ then this rotation is expressed via special unitary matrices $U \in SU\left(2\right)$ as follows : \mathbf{X}^{\prime}\equiv \begin{bmatrix} ...

6

Take your first relation for the 3 Pauli matrices individually: $$\sigma_1(t)=U^\dagger\sigma_1(0)U$$ $$\sigma_1(t)=U^\dagger\sigma_2(0)U$$ $$\sigma_3(t)=U^\dagger\sigma_3(0)U$$ Now you define a "vector" for notational convenience like the OP says in the question. I will choose to rewrite it as a column vector to visually show the relation of the above 3 ...

1

It's a case of bad labeling: the $i$,$j$ labels in Fig.1 and Eqs.(4-5) have different meaning. In addition, subscript 1 was dropped on all $B$'s in Eq.(5). Other than that, it's straightforward algebra: Start by rewriting the final result of Fig.(1) in the familiar operator-product form, expand, and rearrange: $$\overline{\left[ E \cos(B_1\tau) - i {\hat ... 0 The problem here is that you're assuming that |x' \rangle and \langle x| are autokets of g(X) and f(X), respectively. 2 General comments to the question (v1): Any textbook derivation of the correspondence between$$\tag{1} \text{Operator formalism}\qquad \longleftrightarrow \qquad \text{Path integral formalism}$$is just a formal derivation, which discards contributions in the process, cf. e.g. this Phys.SE post. Rather than claiming complete understanding and existence ... 1 I'm sorry I first misinterpreted your braces as anticommutators and not as set inclusions! First absorb the constants into the normalizations of H and B so the (1,1) entry of each is now 1. The common eigenvectors of them both are then the transposes of (1,0,0), (0,1,1), and (0,1,-1); the eigenvalues of each are, under H: 1,-1, -1 ; and under B: 1, 1, -1, ... 1 Operators form a complete set if specifying eigenvalues of all of them already determines a state uniquely (up to normalization and phase). Since they commute, they will have simultaneous eigenspaces. Eigenstate is uniquely determined if all these eigenspaces are one dimensional. 1 I) For the record, here is the operator calculation that OP wants to avoid. The benefit of the calculation is that the operators are not sandwich with any bra/ket representation, and hence we do not have to worry about whether the bra/ket representation is faithful. Let us put \hbar=1 for simplicity. The starting point is the CCR$$ [x^i, ...

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