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4

It is definitely a Kronecker sum. Take the case where there are only two different states $+$ and $-$, then, for example, $$ \hat H =E_+ \hat a^\dagger_+ \hat a_++E_- \hat a^\dagger_- \hat a_- .$$ What does $\hat a_+$ means ? Well, if we label the states with the number of excitations in the states $+$ and $-$ by $|n_+,n_-\rangle$, then we understand $\hat ...


4

The relevant identity is $$\langle x| \hat{p}|\psi\rangle =−i \hbar \frac{d}{dx}\langle x|\psi\rangle\tag{1}$$ which is nothing but the definition of the operator $\hat{p}$. Instead $\frac{d}{dx}|\psi\rangle$ does not make sense as it stands. Because $\frac{d}{dx}$ acts on functions of $x$ whereas $|\psi\rangle$ is a vector in a Hilbert space. Conversely ...


4

The point is that your equation (1) does not make any sense. On the left side, you have operator acting on a vector in some abstract vector space. On the right side you have the position representation of momentum operator acting again on the same abstract vector. Those objects live in different spaces, you cannot just multiply them (actually you can if you ...


3

The state is a vector in the Hilbert space of the Hamiltonian, which gives it a natural basis in terms of the eigenvectors; distinct eigenvalues then exist in distinct (orthogonal) subspaces - for degenerate values the subspaces are larger, but they are still distinct from all others. Clearly this situation gives many advantages in analysis. However, this ...


3

(1) The operator $U(\Lambda,a)$ is a unitary "rotation" in the Hilbert space corresponding to an inhomogeneous Lorentz transformation of the spacetime coordinates. When $U(\Lambda,a)=\exp(iH\tau)$, it is an operator that adjusts the clock forward by $\tau$. Conceptually this is not a physical time evolution of the system. (2) A unitary rotation $U$ in the ...


3

Just one comment to the higgsss answer. Formally from the Wigner theorem we have that if there exist time shift symmetry, for which the scalar product of quantum mechanical rays is conserved, $$ \tag 1 |\langle \psi (t)|\kappa (t)\rangle| = |\langle \psi{'}(t+\tau)| \kappa{'}(t+\tau) \rangle|, $$ then the symmetry transformation acts on $|\psi\rangle$ as ...


3

The correct formula is $$ \mathrm{tr}[A^TB]=\langle m \vert A\otimes B\vert m\rangle\ , $$ so your proof is correct, you're just trying to proof an erroneous formula. (You can easily verify this because with a $\dagger$ the l.h.s. is sesquilinear while the r.h.s. is bilinear.) But I have the feeling this has been asked before. If you have this from ...


3

We can talk about what the notation represents. Terms I introduce will be italicised. For 1, the ket $\left|\psi\right\rangle$ represents the state of a physical system. Quantum mechanics claims these are elements of a vector space. So far, it's all physical. However, everything afterwards will abstract from that. For 2 and 4, the bra ...


3

I think the easiest way to think about these objects is as follows: $|\psi\rangle$ is your physical state Your physical state comes with a machine (its dual) $\langle \psi |$, which when applied to any other physical state $|\phi \rangle$, spits out the overlap $\langle \psi | \phi \rangle$ between your state and $|\phi\rangle$ It also comes with a ...


3

In answer to the title question: no, you can't always decompose an $L_2$ function in terms of only the bound spectrum of hydrogen. This is because there are orthogonal functions to all bound states, which naturally represent the free states of the electron. The quickest examples are of course the Coulomb-wave eigenfunctions $|\chi_{E,l,m}⟩$ of the ...


2

The $a,a^\dagger$ act on the Fock space. If you write a random $\delta(\vec x - \vec y)$ is it neither an element of a Fock space nor an operator on it - the equation doesn't really make sense without further context. However, $\delta$ is just a distribution on functions of spacetime and not operator-valued itself, so the meaning of $a(x)\delta(x-y)$ is ...


2

The trick is the following. Write $$\sum_{a',b,b'} \langle a'|b'\rangle \langle b'|X|b''\rangle\langle b''|a'\rangle = \sum_{b,b'} \langle b'|X|b''\rangle \sum_{a'} \langle b''|a'\rangle\langle a'|b'\rangle .$$ But $$\sum_{a'} |a'\rangle \langle a'|$$ is the identity operator, so the latter sum evaluates to $$\langle b''|b'\rangle.$$


2

Your treatment of the $\hat x$ operator is correct, so I'll focus on the $\langle x|\hat p |E\rangle$ term. The expression $\frac{d}{dx}|E\rangle$ only makes sense if $|E\rangle$ is, for example, some one-parameter family of wave functions indexed by the parameter $x$. This occurs for instance, when computing a geometric phase. Meanwhile, the expression ...


2

OP's ket first equation$^1$ $$\tag{1} \hat{p}|x\rangle ~=~+i\hbar\frac{\partial |x\rangle}{\partial x}$$ is explained in eq. (7) of my Phys.SE answer here. In short, eq. (1) is consistent with the corresponding bra equation $$\langle x |\hat{p} ~=~-i\hbar \frac{\partial \langle x |}{\partial x} $$ via Hermitian conjugation. The bra equation in turn is ...


2

Suppose $\left\{\left|e_i\right\rangle|i\in I\right\}$ is an orthonormal basis of a Hilbert space $\mathcal{H}$, viz. $\left\langle e_i |e_j\right\rangle =\delta_{ij}$. Then the identity operator from $\mathcal{H}$ to $\mathcal{H}$ can be written as an outer product $$\mathbb{I}=\Sigma_{i\in I}\left|e_i\right\rangle\left\langle ...


1

normal ordering is a valid operation provided one can undo it by an appropriate choice of counterterms (of existing couplings or field renormalisations). (How this is done in practice is explained here: http://arxiv.org/abs/1512.02604.)


1

$\newcommand{\real}{\mathbb R}\newcommand{\field}{\mathbb F}\newcommand{\cx}{\mathbb C}\newcommand{\ip}[2]{\left< #1,#2\right>}$We need to dive into mathematics of vector spaces and inner products in order to understand what a vector means and what is it mean to take a scalar product of two vectors. There is a long post ahead so bear with me even ...


1

There is a pattern. If you fix the eigenbasis of $S_z$ you have for any up $+$ or down $-$ eigenstate in $\vec{u}$ direction: $$ \vec{u}=\sin\left(\theta\right) \cos \left(\phi\right)\vec{x}+\sin\left(\theta\right) \sin \left(\phi\right)\vec{y}+\cos\left(\theta\right)\vec{z} $$ then $$ S_u=\sin\left(\theta\right) \cos ...


1

Presumably your vector stands for a three-dimensional quantum state. The Bloch sphere is really only useful for two-dimensional quantum states, as commented by Emilio Pisanty. Theoretically, there is an equivalent description to the Bloch sphere for three dimensional quantum states, but it is not useful for visualization as it is an eight dimensional ...


1

Is measurement just capturing the states of the quantum field at that particular moment? at that particular moment but the measure doesn't depend from this particular moment. There is no cycle nor predetermination. Is measurement just capturing the states of the quantum field it does but it may change the state during the measure. just ...


1

We know that $\psi_{l,m}$ satisfies, for each $l$ and $m$, the equations $$L^2\psi_{l,m}(r,\theta,\phi)=l(l+1)\hbar^2\psi_{l,m}(r,\theta,\phi),$$ $$L_z\psi_{l,m}(r,\theta,\phi)=m\hbar \psi_{l,m}(r,\theta,\phi).$$ But we also know that, by definition, $Y_{l, m}(\theta, \phi)$ satisfy the same equations. Then, unless the eigenvalue equations for $Y_{l,m}$ ...



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