# Tag Info

8

What is your Hilbert space? In $L^2(\mathbb R)$ your eigenfunction would have infinite norm. If you dealt instead with a bounded set $L^2([a,b])$, your operator would not be Hermitian unless you impose suitable boundary conditions to discard boundary terms. These boundary conditions, however, would rule out your candidate eigenvector!

6

If $\cal H$ is a complex Hilbert space, and $A :D(A) \to \cal H$ is linear with $D(A)\subset \cal H$ dense subspace, there is a unique operator, the adjoint $A^\dagger$ of $A$ satisfying (this is its definition) $$\langle A^\dagger \psi| \phi \rangle = \langle \psi | A \phi \rangle\quad \forall \phi \in D(A)\:,\forall \psi \in D(A^\dagger)$$ with: ...

4

Hint: $pq$-order$^1$ your last expression $$2(p^2q^2+q^2p^2)-(pq+qp)^2.$$ $pq$-ordering means commuting all $p$'s to the left and all the $q$'s to the right by using$^2$ the CCR formula $qp = pq +i\hbar{\bf 1}$, possibly repeatedly. (There are shorter ways, but $pq$-ordering is at least a systematic approach.) What remains will be a $c$-number. In fact, the ...

4

The eigenvalue equation $$\tag{1} \hat{x}\psi(x)~=~x_0\psi(x)$$ in the standard Schrödinger position representation $$\tag{2} \hat{x}~=~x, \qquad \hat{p}~=~-i\hbar\frac{\partial}{\partial x},$$ reads $$\tag{3} (x-x_0)\psi(x)~=~0,$$ which has general solution $$\tag{4} \psi(x) ~\propto~ \delta(x-x_0).$$

4

The point is that, if you have a pair of operators $a,a^\dagger$ such that on a common invariant domain $S$, $[a,a^\dagger]= I$, $a^\dagger$ is the restriction of the adjoint of $a$ on $S$ and, in $S$, there is a unique vector $|0\rangle$ with $a|0\rangle=0$, then the closure space spanned by the $|n\rangle$ is isomorphic to $L^2(\mathbb R)$. In that space ...

4

I do not know it this is an answer, since I am not sure to have understood your question. The structure of the equation is formally hyperbolic: $$\frac{\partial^2 \psi}{\partial t^2} - A\psi = S\quad (1)$$ where $\psi =(p,q)^t$. If $A$ were self-adjoint and non-negative (or non positive, changing a sign and inserting a further $i$ in front of $\sqrt{-A}$ as ...

3

Comments to the question: Under the ordering symbol (such as, e.g. normal ordering $:\ldots:$, time ordering $T(\ldots)$, radial ordering ${\cal R}(\ldots)$, etc) all the operators (super)commute, e.g. $$: \hat{A}\hat{B}: ~=~ (-1)^{|\hat{A}||\hat{B}|}: \hat{B}\hat{A}:,$$ even if the (super)commutator $[\hat{A},\hat{B}]\neq 0$ is non-vanishing. Ordering ...

3

I slightly deviate from your notation and use $\phi$ to denote the scalar field as its more standard. Also I should point out that quantum fields are operators and thus under a transformation they get acted on from both the left and the right. The complex scalar field is given by, \phi (x) = \int \frac{ \,d^3p }{ (2\pi)^3 } \frac{1}{ ...

3

The normal ordening is a way to say: ''we throw away the zero-point energy'' (since it becomes infinity and wa say we only look at energy-differences), or to put it in the words of A. Zee: ''Create before you annihalite''. The chronological ordening comes in when you calculate the Feynman propagator (also called the Green's function), which is basically the ...

3

A linear operator $A: D(A) \to {\cal H}$ with $D(A) \subset {\cal H}$ a subspace and ${\cal H}$ a Hilbert space (a normed space could be enough), is said to be bounded if: $$\sup_{\psi \in D(A)\:, ||\psi|| \neq 0} \frac{||A\psi||}{||\psi||} < +\infty\:.$$ In this case the LHS is indicated by $||A||$ and it is called the norm of $A$. Notice that, ...

3

Here's the basic idea behind ladder operators in a bit of generality. Let's say that I have a self-adjoint operator $J$ on the Hilbert space $\mathcal H$ of a given system, and suppose that $\{|m\rangle\}$ were an orthonormal basis for $\mathcal H$ consisting of eigenvectors of $J$, namely \begin{align} J|m\rangle = m|m\rangle. \end{align} Now, suppose ...

2

Linear combinations of Pauli matrices play particularly nicely with diagonalization. The reason for this is that a linear combination of the form $$\vec v\cdot\vec\sigma=\sum_j v_j\sigma_j =\begin{pmatrix}v_z&v_x-iv_y\\ v_x+iv_y&-v_z\end{pmatrix}\tag1$$ represent the density matrix $\rho=\tfrac12(1+\vec v\cdot\vec\sigma)$ of a state at the point ...

2

As the Pauli matrices are hermitian, every linear-combination (with real coefficients) of them is hermitian as well, in particular $(\sigma_x\pm\sigma_y)$. And because every hermitian operator can be diagonalized, the answer to your question is yes. Just write the corresponding 2x2-matrix and try to diagonalize it.

2

The matrix you consider is Hermitian because real linear combination of Hermitian matrices, thus it can be diagonalized (i.e. $S_3$ does exist) for a known general theorem. If a matrix admits eigenvalues it may be non-diagonalizable. Consider the matrix $A$ with the form: 0 1 0 0 The eigenvalues are the complex solutions of $\det(A-\lambda I)=0$. There ...

2

You got the sign on that relation wrong. It should say $\left[q^n, p \right] = i\hbar nq^{n-1}$. A more general relation is $$\left[f(A), B\right] = \left[A, B \right]\frac{\partial f}{\partial A}$$ if $\left[A, \left[A, B \right]\right] =0$. In this case we are okay to use this because $\left[q, p \right] =i\hbar$ is just a number, and that commutes with ...

2

The set of all possible elements of the form $e^{i\alpha}D(x,p)$ with $\alpha, x,p \in \mathbb R$ verifying the commutation relations you wrote in addition to: $$D(x,p)^* = D(-x,-p)\:,\quad D(0,0)=I$$ is a group and it is called Heisenberg group, it is homeomorphic (diffeomorphic) to $U(1) \times \mathbb R^2$ but not isomorphic as a Lie group. It is a real ...

2

One gets there by noting that $\langle x | p \rangle = e^{i p x/\hbar}$ is a plane wave, and you have to throw on the test wave function to talk about the derivative operation. So, you are worried about $$\langle x| P | \Psi \rangle = \int dp ~p~ e^{i p x/\hbar} ~\langle p | \Psi \rangle$$ From there you note that you can get the ...

1

You just made some math mistakes. You made a mistake when you did $Q = h\int_A kr$. You got $Q = h\pi k r^2$, but you should have gotten $Q = \frac{2}{3} h\pi k r^3$. Notice how this second expression has units of charge while the first one doesn't. Another mistake you make is that you say $\frac{1}{r} \frac{\partial rE(r)}{\partial r} = \frac{1}{r} ... 1 We can realize the displacement operator as $$\tag{1}\hat{D}(x,p)~=~e^{x\hat{P}+p\hat{X}},$$ where the elements$\hat{X}$,$\hat{P}$and${\bf 1}$generates the Heisenberg algebra $$\tag{2} [\hat{X},\hat{P}]=i{\bf 1}.$$ These elements can be realizes as differential operators in the Schrödinger representation. (See also the Stone-von Neumann theorem.) ... 1 The latter description is correct (as is described in Sakurai, Gasiorowicz, Griffiths, and probably some other books that I don't own). What it is saying is that the inner product between$|x\rangle$and$|x_0\rangle$is either 0 if$x\neq x_0$or 1 if$x=x_0$. That is, the states are orthogonal. The momentum space description $$\langle ... 1 Since [\hat{p},\hat{T}]=0 and \hat{T} is unitary (\hat{T}\hat{T}^{\dagger}=\hat{T}^{\dagger}\hat{T}= \mathbf{1}) we have that \hat{T}^{\dagger} \hat{p} \hat{T}= \hat{T}^{\dagger}\hat{T}\hat{p}=\hat{p}. Edit after a question in the comments: [\hat{T}, \hat{p}]=0 holds. For simplicity let us use the \hbar=1 units. We know that ... 1 This is much to do with the possible eigenvalues of the operators. Normal operators on a Hilbert space are closely analogous to complex numbers, with the adjoint taking the role of the conjugate; these relations are typically inherited directly to the operator's eigenvalues. Thus, if a linear operator L has an eigenfunction f with eigenvalue \lambda, ... 1 Normal-ordered non-singular terms in the OPE are still there in principle, but they are sometimes omitted (and therefore only implicitly implied) in the notation. This is because many important physical quantities only depend on the singular terms of the OPE. Btw: Also note that many authors don't write the radial ordering symbol \cal R explicitly. This ... 1 Are p and q the standard momentum and position operators in L^2(\mathbb R)? If the answer is positive, then:$$K_\pm := K_1 \pm iK_2 = \frac{1}{2}\left(\frac{1}{\sqrt{2}}(p\pm iq) \right)^2\:.$$In other words, introducing the standard operators$a = \frac{1}{\sqrt{2}}(p- iq) $and$a^\dagger = \frac{1}{\sqrt{2}}(p+ iq) $for the harmonic oscillator: ... 1$[q_r q_s p_r , q_s p_r q_s]=q_r [q_s p_r , q_s p_r q_s ] + [q_r q_s , q_s p_r q_s]p_r=q_r q_s [p_r , q_s p_r q_s ] + q_r[q_s , q_s p_r q_s]p_r + q_r [q_s , q_s p_r q_s ]p_r + [q_r , q_s p_r q_s]q_s p_r$Now only the last term is non-vanish because the others three have inside the commutators only operators that commutes one to each other. So:$[q_r q_s ...

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