# Tag Info

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You are making confusion between operators and their representations on the position basis. Your hypotheses 1) and 2) are wrong (more ill-interpreted): as you can easily notice the left hand sides contain an element in the Hilbert space whereas the right hand sides contain numbers (functions evaluated in a point $x$, therefore a number). To answer your ...

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It you are working with second quantization (quantum field theory), then what you wrote under "1." and "2." is incorrect. Your physical degrees of freedom are not coordinates $x$, but rather field variables $\phi$. So, $$\left| 0 \right> \sim \exp \left\{ -\frac{1}{2} \int d^3 x \phi(x)^2 \right\}$$ This object should be called wave functional ...

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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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Equation (2.4.6): $T(z)X^\mu(0)\sim \frac{1}{z}\partial X^\mu(0)$ means that the RHS is the most singular term of the LHS. $T(z) = -\frac{1}{\alpha'} :\partial X^{\mu} \partial X_{\mu}:\tag{2.4.4}$ So \begin{align*} T(z)X^{\mu}(0) & =-\frac{1}{\alpha'}:\partial X^{\nu}(z)\partial X_{\nu}(z):X^{\mu}(0)\\ & =-\frac{2:\partial ...

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Everything follows is valid at least when the various operators appearing below are defined on the dense subspace finitely spanned by the standard vectors $|n\rangle$, $n=0,1,2,\ldots.$ From $[a,a^\dagger]=I$ and $[A,BC]= B[A,C]+[A,B]C$, by induction, one easily finds that $$[a,(a^\dagger)^n] = n(a^\dagger)^{n-1}\:.$$ If $\alpha_k \in \mathbb C$ and ...

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A time reversal operator is an anti-unitary operator, which can be expressed as: $\mathcal{T}=UK$ where $K$ denotes complex conjugate and $U$ is a unitary operator. In case of spinless particles, $U$ is chosen to be Identity. Thus $\mathcal{T}=K$. If the system has $\mathcal{T}$-reversal symmetry: $$KH\psi=HK\psi$$ which leads to: $$... 4 By definition, a value is in the continuous spectrum of A if it not an eigenvalue, but the range of A-\lambda I is a proper dense subset of the Hilbert space. There is nothing in this definition distinguishing separable spaces, or precluding operators in them from having continuous spectrum, and indeed some do. There is an equivalent definition in terms ... 0 Sakurai's argument leading to eq. 2.3.13 depends on the property of the operator N and the related labelling of its eigenstates, which is explicitly given as:$$N~|n\rangle = n~|n\rangle \tag{2.3.7}.$$Of course this is meant to hold for any particular integer value of n \ge 0; therefore, likewise, for any integer k = n + 1 \ge 1:$$N~|k - 1\rangle ...

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It may be confusing that there are more that one Hilbert spaces, but it is something that we also see in describing a particle in space. Thinking of the energy eigenstaates, in 1D free space you will have a superposition of an uncountable number of momentum eigenstates ($k$). However if you have a infinite well of width $a$, the energy eigenstates are now ...

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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).

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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 ...

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If you read the next paragraph, it seems that Dirac means that the eigenvalues are non-degenerate, i.e. for a set of simultaneous eigenvalues $\xi_1^\prime, \xi_2^\prime\dots$ there is exactly one corresponding eigenbra. There are no two distinct bras which are eigenbras of all the operators $\xi_1, \xi_2\dots$ and have the same eigenvalues for all of them. ...

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Hint: Think of way to write the Hamiltonian as $$\hat{H}~=~ -\frac{\hbar^2}{2m}\frac{d^2}{dx^2} +V(x)~=~\hat{B}^{\dagger}\hat{B} ~\geq~0$$ for some first-order differential operator $$\hat{B}~=~a(x)\frac{d}{dx} +b(x) ,$$ with suitable functions $a(x)$ and $b(x)$. Here the potential $V(x)$ is given by formula (1). Can you see what the operator ...

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As far as my experience goes (theoretical nuclear physics), they are pretty much equivalent. To make this clearer, I'll try to give an intuitive example for bosons: $$\langle cc{}^\dagger\rangle = cc{}^\dagger-:cc{}^\dagger:$$ which becomes $$\langle cc{}^\dagger\rangle = cc{}^\dagger-c{}^\dagger c = \left[c,c{}^\dagger\right]$$ so in this case $\langle ... 0 After thinking about it, as long as the original eigenvalues are non-degenerate it should be possible to have the new Hamiltonian be represented by a differential equation of arbitrarily high order. The key is that the projection operators$P_n$onto the eigenfunctions exist in the algebra generated by the original Hamiltonian$\hat{H_0}$. For instance say ... 0 If the electron is confined to the$x-y$-plane, it's$z$-position is fixed, i.e. certain, and hence the$z$-momentum infinitely uncertain by the uncertainty relation. That paragraph is trying to say that, if the magnitude of$\vec L$is larger than$L_z$, then$\vec L$is not fixed, and if angular momentum is not fixed, i.e. conserved, then the motion does ... 0 If the particle is known to be in the$xy$-plane, then$\Delta z = 0$, and so (by the uncertainty principle)$\Delta p_z = \infty$. Roughly speaking, this means that there's a good chance that the particle's momentum would be greater than "escape momentum", and thus that it could escape from the hydrogen atom. (I'm kind of dubious about this argument ... 0 There is a derivation in Fetter and Walecka, as well as another treatment in Sakurai. I haven't looked at these in a while and they may or may not provide the mental satisfaction you are looking for, but it is worth looking at these two treatments at any rate. 0 Your second option is correct. It is a simultaneous eigenket of all three commuting operators. And that is how it should be interpreted. To be explicit, the equation $$\hat{\mathbf{x}}\mid\mathbf{x'}\rangle = \mathbf{x'}\mid\mathbf{x'}\rangle~?$$ Is a triple of equations$\hat{x}\mid \mathbf x'\rangle = x'\mid \mathbf x'\rangle\hat{y}\mid \mathbf ...

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$\newcommand{\ket}[1]{\lvert #1 \rangle}$The Schrödinger equation does not say what you claim it does. The time-independent Schrödinger equation is an eigenvalue equation for the Hamiltonian operator $$H \ket{\psi_E} = E \ket{\psi_E}$$ where solving for $\ket{\psi_E}$ for a concrete $H$ gives us the occuring eigenstates of $H$. It does not say that these ...

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$\newcommand{\ket}[1]{\left| #1 \right>}$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 ...

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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 ...

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Note that you only can identify the numbers with the operators when you hold a plane wave function. For an arbitrary wave functions, you can't identify the operator $(\hbar/i)(\partial / \partial x)$ with some number $p$, but you can hold the operator to have some significant meaning related to this numbers. If you apply $$\langle P\rangle = \int dx \, ... 7 Not equal, but equivalent, in the sense that they have the same effect on the wavefunction in question. More precisely, the book is using a slight abuse of terminology. Taking momentum as an example, it's not really the case that the dynamical quantity of momentum is equivalent to the operator \frac{\hbar}{i}\frac{\partial}{\partial x}, because a number ... 2 As argued by von Neumann, the measuring process has many properties that resemble those found in the theory of operator algebras. For instance, if you have an instrument, you can measure something, say the length of a table, to get a certain value x within experimental errors. What you can now do is relabel the ticks of your instrument according to a ... 8 The reason operators correspond to measured values has to do with what happens when you connect a measurement apparatus to the system under observation. Suppose the Hamiltonian of the system by itself is H_S and the Hamiltonian of the measurement apparatus by itself is H_M. When M is physically connected to S, we get a additional "interaction" term ... 0 You will have to settle for momentum since speed is not a quantum mechanical observable (because \dot{x} is not a classical observable on the phase space, which are functions of x and p, but a function of a classical trajectory (x(t),p(t)), so canonical quantization does not produce a "speed" observable). The probability for a certain momentum, ... 1 If you have an observable, that is like a symmetric matrix. So its eigenvectors span the whole space so you can form a basis for the whole space. So the span of the eigenvectors really is the whole space. What about eigenspaces? You did have to use all the eigenvectors to get the whole space, not just ones that share the same eigenvalue. If you only use ... 2 Short answer: yes A map A: V \to V on a \mathbb{K}-vectorspace V is linear if \forall v_1, v_2 \in V and \forall \lambda \in \mathbb{K} we have:$$A(\lambda v_1 + v_2)= \lambda A(v_1)+A(v_2)$$If we have two such maps A,B then [A,B]=AB-BA is again linear because, well AB is linear:$$(AB)(\lambda v_1+v_2)=A(B(\lambda v_1 + v_2))=A(\lambda ...

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By definition $$\alpha^*=\bar{\alpha}$$ and is called a complex conjugate so for the real part, the complex conjugate is itself while the imaginary part takes on the opposite sign. To prove that $$\langle A|=\langle B|\bar{\alpha}$$ first define this equation to be the complex transpose of $$|A\rangle=\alpha|B\rangle$$ For the complex number alpha, the ...

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Observe that $$A = \frac{1}{2}(A + A^\dagger) + \frac{1}{2}(A - A^\dagger)$$ where the summands are Hermitian and anti-Hermitian as desired.

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Step 1: Write the function $\phi(p,t)$ in terms of its Fourier components, i.e.: $\phi(p,t) = \int dx \phi(x,t) e^{i p x}$ Step 2: Notice that the $\nabla$ operator acts now only on the exponential factor inside the integral. Doing the derivative results in: $\int dx (-x^2) \phi(x,t) e^{i p x}$ Step 3: Multiply the equation by by a factor of $e^{-i p r}$ ...

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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 $|\phi\rangle$.

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Elaborating on ACruiosMind's comment, assume that the matrices $A$ and $B$ are defined the following way: $$A=\begin{pmatrix} 1 & 2 \\ 5 & 4 \end{pmatrix} \quad \text{and} \quad B = \begin{pmatrix} 1 & 1 \\ -1 & -1 \end{pmatrix}$$ Notice that the eigenvectors of $A$ are $$\begin{pmatrix} 1 \\ 5/2 \end{pmatrix} \quad \text {and} \quad ... 0 Although  [ \hat{L_{x}}, \hat{L_{y}}] \phi_{l, m_{l}} = 0, \phi_{l, m_{l}}  is neither \hat{L_{x}}'s nor  \hat{L_{y}}'s eigenstate. 1 Commuting matrices We take two Hermitian matrices A and B that commute. Since they are Hermitian and act on the same Hilbert space they have a common (orthonormal) basis of eigenvectors. In general all Hermitian matrices have the same amount of lineary independent eigenvectors since they all have a orthonormal basis of eigenvectors. Commuting observables ... 4 As noted by Sebastian, [Q,H] will be anti-Hermitian and therefore generally not an observable (except in the trivial case). However i[Q,H] is an important observable. This corresponds to the classical Poisson bracket which can be see in the following formula,$$ \frac{d \langle Q\rangle}{d t} = \frac{i}{\hbar} \langle [H,Q]\rangle + \langle ...

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Do $A$ and $B$ have the same amount of eigenvalues? They don't have to, $S_z$ and the identity commute. The former has two eigenvalues. The latter has one. Other examples exist too, such as the operator, $H=p^2/2m+e/r$ which has an infinite number of eigenvalues but can commute with operators with a finite number of eigenvalues such as both operators ...

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$[A, B]$ for two observables $A$ and $B$ is an observable if, and only if, $A$ and $B$ commute. Proof: $$[A, B]^\dagger = (AB)^\dagger - (BA)^\dagger = B^\dagger A^\dagger - A^\dagger B^\dagger = BA - AB = -[A, B].$$ Note: An observable is any Hermitian operator. The commutator of two Hermitian operators is anti-Hermitian, as the proof shows. $0$ is an ...

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I'm not quite sure what your specific question is, so I'll try to better explain what Griffiths is doing in his book. In first-order perturbation theory, the Zeeman correction to the energy is: \begin{align*} E_{Z}^{1} & = \langle n \; l \; j \; m_{j} \vert H_{Z}^{\prime} \vert n \; l \; j \; m_{j} \rangle \\ & = \langle n \; l \; j \; m_{j} ... 1 Yes, there is a relation between both things. Letg(t)=\exp(tX), \qquad g(0)=e$$Be a curve on G, such that \gamma_a(t)=g(t)\cdot a Then, by deffinition, for every smooth function f:G\to\mathbb{R} this curve satisfies the differential equation$$\frac{d}{dt}(f\circ g)(t)=X(f\circ g)(t)$$or directly$$\frac{d}{dt}g(t)=X(g(t)) In an analogous way, ...

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