New answers tagged

2

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 > ...


2

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 ...


0

Given two operators $A$ and $B$, the pairing of the two operators forms another operator, either $Q=AB$ or $S=BA$. Given the commutator $$[A,B]=j$$ we can write $$Q-S=j$$. Now you can use the standard definition of expectation values to determine which form to use because $$<Q>-<S>=<j>.$$ Edit: The standard definition of the expectation ...


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 : \begin{equation} \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 ...


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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, ...


0

and as we represent the wave function in general as $e^{ikx}$ This isn't true. In general, we represent the wave function as $$\psi(\mathbf{x}) = \langle\mathbf{x}|\psi\rangle$$ Only in the case that $|\psi\rangle = |\mathbf{p}\rangle$ (a momentum eigenstate) do we have $$\psi(\mathbf{x}) = \langle\mathbf{x}|\mathbf{p}\rangle = ...


2

A wave function is an abstract mathematical function which could completely describe about the system under consideration. We define a wave function such that we could derive whatever information from it, provided that will not affect the state of the system. The wave function is not any operator. It's simply a function of position and time. Any ...


4

This is because $H'=UHU^{-1}$ for a certain unitary operator $U$, therefore $\psi$ is an eigenvector of $H$ with an eigenvalue if an only if $U\psi$ is eigenvector of $H'$ with the same eigenvalue. Thus the two operators have the same point spectrum. $U = e^{i \lambda X/\hbar}$. From $[X,P]= i \hbar I$ one finds $$e^{-i \lambda X/\hbar}Pe^{i \lambda ...


0

From the equation $H$ is a linear function of $H_0$. In that case, the eigen values of $H_0$ are eigen values of $H$ also. That's why the eigen values remain invariant under such a transformation.


0

So after much deliberation, I think I have found the answer: $$ \hat{a }^\dagger \hat{a} = \hat{N} = \hat{a }^\dagger[\hat{a},\hat{a }^\dagger]\hat{a}$$ $$ = \hat{a }^\dagger(\hat{a} \hat{a }^\dagger - \hat{a }^\dagger\hat{a})\hat{a}$$ as $\hat{a}^2 = 0$, we have: $$ \hat{a }^\dagger \hat{a} = \hat{N} = \hat{a }^\dagger \hat{a}\hat{a }^\dagger \hat{a} = ...


3

When one writes $A=c$, where $A$ is an operator and $c$ is a number, it is implicit that the r.h.s. actually denotes $c$ times the identity.


2

The precise statement of "self-adjoint operators generate continuous unitary symmetries" is Stone's theorem. It guarantees that there is a bijection between self-adjoint operators $O$ on a Hilbert space and unitary strongly continuous one-parameter groups $U(t)$ that is given by $O\mapsto \mathrm{e}^{\mathrm{i}tO}$. The definition of the exponential for an ...


1

Expectation values of constants or numbers are just those constants or numbers. The expectation value of the anti commutator of $\hat x$ and $\hat p$, that is, $\langle\{\hat x,\ \hat p\}\rangle$, for the Harmonic Oscillator, or coherent states of the Harmonic Oscillator, is equal to $0$. $$\langle\{\hat x,\ \hat p\}\rangle = \langle\hat x \hat p + \hat p ...


1

To quantize just means to impose this commutation relation: $$ [x_i , p_j ] = i \hbar \delta_{ij}.$$ By the way, remember that a function of position $f(x)$ may not commute with the momentum operator $p$. $A_i = A_i(x)$. Do $A_i$ and $p_j$ necessarily commute?


0

If you are quantizing the system "particle in external electromagnetic field", then no, the electromagnetic fields will not become operators themselves. However, since they are functions in $x$, the vector potential $A$ actually becomes an operator $A(\hat{x})$. It is treated as a fixed addition to the Hamiltonian, it is not a dynamical variable itself. For ...


0

When adding angular momentum $S$ and $L$ to get $J$, $J$ can range from absolute value of $|L-S|$ to $L+S$ in integer steps. Then once you have $J$ you can figure the allowed values of $m_j$ just like you do for $S$ and $m_s$


2

It's simply the most general kind of interaction Hamiltonian you can write down in this simplified two-level system. On the 2D Hilbert space spanned by $\lvert R\rangle,\lvert L \rangle$, the most general linear operator is written as $$ A = a_\text{RR}\lvert R\rangle\langle R\rvert + a_\text{RL}\lvert R\rangle\langle L \rvert + a_\text{LR}\lvert ...


2

The answer is given by Prahar in his comments : \begin{equation} {\bf T} \cdot {\bf T}^\dagger= {\rm T}_{1}{\rm T}^\dagger_{1}+{\rm T}_{2}{\rm T}^\dagger_{2}+{\rm T}_{3}{\rm T}^\dagger_{3} \tag{01} \end{equation} For $k=1,2,3$ \begin{equation} {\rm T}_{k}{\rm T}^\dagger_{k}=\dfrac{1}{2\hbar}\left(\sqrt{m\omega}\ {\rm R}_{k}-\dfrac{i}{\sqrt{m\omega}} {\rm ...


3

$ \def\ee{\mathrm{e}} \def\ii{\mathrm{i}} \def\dd{\mathrm{d}} $ There are many ways to perform this calculation. Perhaps the simplest is to consider the object $a(t) = \ee^{\ii \omega_c t a^\dagger a } a \ee^{-\ii \omega_c t a^\dagger a }$ as the solution of the differential equation $$\dot{a}(t) = -\ii\omega_c a(t).$$ You can show that $a(t)$ satisfies ...


7

This is a supplement to freude's correct answer: Hamiltonian is the infinitesimal generator of time translation defined as $$\mathrm{\hat{U}}(\mathrm dt)= 1- \frac{i}{\hbar} \mathrm{\hat{H}}(t)\mathrm dt\;.$$ Time-Evolution Operator: Let the system be at $|\phi\rangle\;.$ Now, let's wait for some time..... What is the probability amplitude of finding ...



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