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## New answers tagged operators

2

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 ...

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 ...

0

may be you can think this way. The wavefunction $|x\rangle=e^{i(kx-\omega t)}$ How do we extract the momentum $\hbar k$ out of it by some operator ? $$\frac{\partial}{\partial x}|x\rangle=ik|x\rangle$$ to get $\hbar k$, you need to either multiply the operator with $-i\hbar$ or with $\frac{\hbar}{i}$. So the momentum operator is $$... 2 OP's ket first equation^1$$\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} $$via Hermitian conjugation. The bra equation in turn is ... 3 Just one comment to the higgsss answer. Formally from the Wigner theorem we have that if there exist time shift symmetry, for which the scalar product of quantum mechanical rays is conserved,$$ \tag 1 |\langle \psi (t)|\kappa (t)\rangle| = |\langle \psi{'}(t+\tau)| \kappa{'}(t+\tau) \rangle|, $$then the symmetry transformation acts on |\psi\rangle as ... 3 (1) The operator U(\Lambda,a) is a unitary "rotation" in the Hilbert space corresponding to an inhomogeneous Lorentz transformation of the spacetime coordinates. When U(\Lambda,a)=\exp(iH\tau), it is an operator that adjusts the clock forward by \tau. Conceptually this is not a physical time evolution of the system. (2) A unitary rotation U in the ... 2 This subtlety is related to the fact that the momentum operator \hat{P} (unlike the Hamiltonian \hat{H}=\frac{\hat{P}^2}{2m}) has no eigenfunctions compatible with the Dirichlet boundary conditions, and \hat{P} is not a self-adjoint operator. This is essentially Example 4 in F. Gieres, Mathematical surprises and Dirac’s formalism in quantum mechanics, ... 3 The action of the boost on momentum states specifies the matrix elements of T_V in momentum representation, it does not "change the momentum representation of the state". In other words, from$$ T_V\left|p_1,p_2,..,p_N\right\rangle=\left|p_1+m_1V,p_2+m_2V,..,p_N+m_NV\right\rangle $$it just follows that$$ \langle p'_1,p'_2,..,p'_N | T_V ...

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I would say the answer is rather simple. If you do various experiments to measure the position (or momentum) of a particle, then QM only gives you consistent and accurate predictions when the mathematical symbols $x$ and $p$ are associated with the measured position or momentum. However, position and momentum are not the only canonically conjugate operators ...

3

The answer may be deeper than you expect. The operator $K$ that the OP defined --- here it will be called $-i\nabla_x$ --- is the generator of space translations. And the operator $X$ --- here called $x$ --- is the generator of momentum translations. If you exponentiate them, to form the corresponding one-parameter unitary groups of translation, you get ...

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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 ...

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 ...

3

$\frac{\partial\hat{Q}}{\partial t}$ denotes the partial derivative of time, which is nonvanishing only when $\hat{Q}$ manifestly depends on time. Every operator $\hat{Q}$ can be time dependent in an implicit way such as $\hat{x}\hat{p}$ which can be time dependent when $\hat{x}$ or $\hat{p}$ depends on time.

1

Define $D=\{(x,y)\in\mathbb R^2:\|(x,y)\|\leq R\}$ as the disk of interest. There are two spaces of interest here: the space of square-integrable functions on $D$, $L_2(D)$, and the space of such functions with Dirichlet boundary conditions, $\mathcal H=\{\psi\in L_2(D):\psi(p)=0\:\forall p\in \partial D\}$. You're interested in the hamiltonian ...

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 ... 9 What you call an operator theory is usually called the Heisenberg picture of quantum mechanics. What you call a wave function theory is usually called the Schroedinger picture of quantum mechanics. It is well-known that for every quantum mechanical model, the Heisenberg picture and the Schroedinger picture are fully equivalent through a dual description, ... 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 $$\langle b''|b'\rangle.$$ 1 Two things are required to prove this. First, you haven't really defined your state$|p\rangle$in its entirety since you haven't defined what normalization you are using for this state. In the formula above, it seems to me that you are using the normalization $$\langle {\bf p} | {\bf p}' \rangle = \frac{2 E_p }{ 2\pi } \delta^3 ( {\bf p} - {\bf p'} )$$ ... 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.) 2 No. There are multiple problem with your idea that$\langle q_2;t_2\vert q_1;t_1\rangle$represents "the probability amplitude for $$Q(t_2)\lvert q_1;t_1\rangle = q_2\lvert q_1;t_1\rangle$$ being true": You have no reason at all to believe that$\lvert q_1;t_1\rangle$will be an eigenstate of$Q(t_2)$, so...this will probably never hold. There are no ... 0 Your initial identification, $$J=(J_1\otimes I)+(I\otimes J_2),$$ is correct. However, it's incorrect to change this sum into a product: $$(J_1\otimes I)+(I\otimes J_2)\neq J_1\otimes J_2.$$ (If nothing else, you want$J_1+J_2$to double if you double both$J_1$and$J_2$, whereas$J_1\otimes J_2$quadruples under that transformation.) Then, if you want to ... 0 If you have$R|r\rangle=r|r\rangle$then you have, where$I$is the identity operator,$|r\rangle=I|r\rangle=R^{-1}R|r\rangle=rR^{-1}|r\rangle$and you immediately have$R^{-1}|r\rangle=\frac{1}{r}|r\rangle$0 Since identity acting on any eigenvector of$R$gives the eigenvector itself, so the result you wrote down for the operator$1/R\$ stands true. Nothing extra needs to be defined. Also you can now go to higher powers by the repeated operations on the eigenvectors by the operators.

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