New answers tagged

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 ...


0

It's hard to picture the things you want because you're working in an infinite-dimensional space, and I don't know anybody who can intuitively 'see' in that kind of space. Instead, let's consider a finite-dimensional space: the polarization of a massless spin 1 particle traveling along the $z$ axis. The polarization state is simply described by a unit ...


0

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 ...


0

$\newcommand{\real}{\mathbb R}\newcommand{\field}{\mathbb F}\newcommand{\cx}{\mathbb C}\newcommand{\ip}[2]{\left< #1,#2\right>}$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

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 ...


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 ...


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 ...


3

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

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 ...


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 ...


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 ...


0

Okay, so there's two things to understand here: quantum field theory and quantum measurement. They are not very well-integrated at present; the best work in quantum measurement is done with "vanilla" quantum mechanics; the best work with quantum field theory takes a completely different tack from this and deals with Lagrangian densities and other such crazy ...


0

There are two kinds of operators which have an intuitive meaning when you apply them to a state: Evolution operators like $e^{-i\hat H\Delta t/\hbar}$ simply take the state at time $t$ and give you the state at $t+\Delta t$ Projectors tell you what the state will be after you take a measurement. For example, supose you measure an observable $\hat A$ on ...


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 ...


0

Such a vector describes the quantum state of a spin-1 particle on a line, or any other particle with a position degree of freedom and 3 internal states. To start with, you can expand wavefunctions in a basis. E.g., if you have a wavefunction $\vert\psi\rangle$ which depends on position, you can expand it in the position basis $\vert x \rangle_p$, i.e., ...


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 $$\langle b''|b'\rangle.$$


0

A state $|\psi\rangle$ in quantum mechanics is a piece of information about a system that can be copied. $|\psi\rangle\langle\psi|$ and $\langle\psi|$ are both best understood as alternate representations of that same state. And $\langle\psi|\psi\rangle$ is a complex number such that $|\langle\psi|\psi\rangle|^2$ is the probability of the state ...


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

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 ...


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.)


0

Dealing with rigged Hilbert spaces, the space of bras and the space of kets are not isomorphic, though the space of ket vectors is identified with a subspace of bra vectors. Let us focus on the simplest case: ${\cal H} = L^2(\mathbb R)$. Here, the space of bra vectors is the space of Schwartz distributions ${\cal S}'(\mathbb R) \supset {\cal H}$, whereas ...


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.



Top 50 recent answers are included