# Tag Info

11

The word "span" is meant in a slightly different way than in elementary linear algebra, where (as you say) it wouldn't make sense for a set of vectors to span a space without actually being in it. As mentioned in the comment by Charlie, this requires the notion of rigged Hilbert spaces to actually formalize. If you want the end-user result, it ...

8

That is because the adjoint $A^*:D(A^*) \to H$ of an operator $A: D(A) \to H$ exists if $D(A)$ is dense in $H$. Next, since $A=A^*$ if $A$ is selfadjoint, we have that $D(A)=D(A^*)$ is dense. The requirement $D(A)$ dense for defining $A^*$ has the following reason. By definition we have both $$D(A^*) := \{x \in H \:|\: \langle x| Ay \rangle = \langle z_x| y\... 5 Evolution of a quantum state is unitary, that is it can be wrtitten as$$ |\psi(t)\rangle = U(t)|\psi(0)\rangle $$where U(t) is an unitary operator. Hamiltonian is the generator of this evolution, that is the operator U(t) satisfies the equation$$ \frac{dU(t)}{dt} = -\frac{i}{\hbar} H(t) U(t)$$from which we can calculate$$ H(t) = i\hbar \frac{dU(t)}{...

5

I hope this will help you to understand some of the statements. I will slightly modify the notation. We will denote the adjoint of an operator $A$ by $A^{\dagger}$ (in your post this corresponds to $\overline{A}$). If $A=A^{\dagger}$, then we call $A$ hermitian (='real'). We further say that two operators $A$ and $B$ commute if $[A,B]\equiv AB-BA = 0$. In ...

5

Let me give you the most general definition which holds for completely generic QFTs (including those which do not have a Lagrangian formulation). The specific case of QED or QCD is described in the answer by @CosmasZachos. A current operator is any spin-1 operator which satisfies the following Ward identity $$\nabla^\mu \langle J_\mu(x) O_1(x_1) \cdots O_n(... 5 Hermitian is an adjective used to describe an operator which is equal to its Hermitian conjugate. Hermitian conjugate (sometimes also called Hermitian adjoint) is a noun referring to the generalisation of the conjugate transpose of a matrix. It doesn't really make sense to say that a particular operator is a Hermitian conjugate without any context. In your ... 5 The Hermitian Conjugate or Hermitian Transpose of an operator \hat{O} is defined as \hat{O}^\dagger. As you stated in your question an operator \hat{Q} is Hermitian iff \hat{Q}=\hat{Q}^\dagger, I know the terminology can be confusing. An operator is Hermitian if it is equal to its Hermitian Conjugate. Now to the differentiation operator: I assume ... 4 Generically, it is the coefficient of the term linear in the gauge field in a gauge-invariant action, so it is the "matter stuff" a gauge field propagator abuts to in a Feynman diagram. Classically, it corresponds to a conserved global-invariance Noether current associated to a conserved charge. In the gauge theory, it is the linchpin of gauge-... 4 There is already a good answer from J. Murray. Here we merely want to answer OP's last question. How can vectors not be a part of a vector space but still span it? Answer: E.g. a basis (\vec{e}_1,\ldots,\vec{e}_n) of a finite-dimensional vector space V spans by definition V and therefore a subspace U\subset V even if none of the basis vectors ... 4 Let me first state that functional determinants are not the same as determinants. So point A is wrong. About B, there is no single best way to do it, it is generally a hard task. Let me know elaborate. So a functional determinant is the determinant of a differential operators, that is the determinant of of a linear functional in a function space, that means ... 4 If an expectation value is imaginary, it means the expectation value is imaginary. An expectation value is just a statistical attribute of the distribution of observables. Hermitian operators have real observables, in the form of real eigenvalues; and correspondingly the expected value of those operators must be real as well. Since every actual observable is ... 3 I know that a Hamiltonian is a self-adjoint operator which describes the total energy of a system, and its eigenvalues refer to the possible energy levels of that system. If you knew that, you wouldn't be asking the following question: Is there a unique Hamiltonian that characterizes the state  |ψ(t_1)⟩ , or the operation applied? If not, what exactly ... 3 Not an answer, but an extended comment on your basically sound approach, since the comment format does not permit such extended comments. The group involved is the oscillator group, and the 3d rep you found is a faithful one, so any group relation for it will also hold for the abstract group in general, so, all representations! I will call your central ... 3 The difference is likely a result of viewing transformations as being passive, as opposed to active. The difference can be a little subtle. In an active transformation, we actually move the state to a new position. In a passive transformation, we instead move the coordinate system in the opposite direction. In this case, there are two possible ways to define ... 3 As Jakob commented, to prove identities of that kind it is often good to go back to the definition of the adjoint operator as arising from an inner product. Given an inner product (\cdot,\cdot) and an operator \hat{A}, one defines the adjoint operator \hat{A}^\dagger to be the operator that satisfies$$(v,\hat{A}w) = (\hat{A}^\dagger v,w)$$for all ... 2 You cannot do that because it is not correct! What you can do instead is expressing the operator \hat{A} in the complete basis of the position variable x:$$\langle x|\hat A|a_k\rangle= \int dx^\prime \langle x|\hat A|x^\prime\rangle \langle x^\prime|a_k\rangle = \int dx^\prime A(x,x^\prime) u_k(x^\prime)$$where A(x,x^\prime) := \langle x|\hat A|x^\... 2 As @Qmechanic points out in the comment above, they are both correct. This is because they are both equivalent. I'm not an expert at the rigorous math, but the way I think about it is that "functions" like the Dirac Delta are actually distributions, meaning that they only make sense inside of an integral of the form:$$\int_{a}^b \text{d}x\,\,f(x)\...

2

You appear to have written down your matrix elements ambiguously. Recall $$|\psi\rangle= \tfrac{1}{\sqrt{2}} (|m\rangle +|n\rangle), \\ x_{mn}= \langle m|\hat x |n\rangle= \langle m|( a^\dagger + a )|n\rangle=\sqrt{n+1} ~\delta_{m,n+1}+ \sqrt{n} ~\delta_{m,n-1}.$$ It is then evident that $\langle \psi|\hat x|\psi \rangle = (x_{10}+x_{01})/2=1$, for your ...

2

Yes. Recall the following facts: The definition of the derivative of an operator which depends on some parameter is given by $$\frac{d}{dx}A(x) = \lim_{\epsilon\rightarrow 0}\frac{1}{\epsilon}[A(x+\epsilon)-A(x)]$$ For all $|\psi\rangle$ in the vector space, the limit of a family of operators $A(x)$ is defined by $$\left[\lim_{x\rightarrow x_0} A(x)\right]|... 2 I'll elaborate on the answer, since it took me some effort and I'm glad to share all the steps. Applying the Cauchy-Schwarz inequality to |\langle \psi| U^\dagger M |\Delta\rangle |  it follows that:$$|\langle \psi| U^\dagger M |\Delta\rangle | \leq ||\psi|| \: ||U^\dagger M \Delta || $$Note that angle brackets and vertical bar of the bra-ket notation ... 2$$|\langle \psi| AB \phi\rangle | \leq ||\psi|| \: ||AB \phi|| \leq ||\psi||\: ||AB|| \: ||\phi|| \leq ||\psi|| \: ||A||\:||B||\: ||\phi||$$In our case ||\psi||=1 and ||A||, ||B|| \leq 1. because one of A and B is unitary and the other is part of a POVM. 2 I'll sum up the previous comments. Remeber that, ignoring the scaling factor \hbar/2 for S_x, the following equation holds:$$\left({S_x}\right)^{2} = \begin{pmatrix} 0&1 \\ 1&0\end{pmatrix}\begin{pmatrix} 0&1 \\ 1&0\end{pmatrix} = \begin{pmatrix} 1&0 \\ 0&1\end{pmatrix} = \mathbb{I}\,$$Then expand the exponential operator: \... 2 Let me begin by noting that S and \sigma should generically not be confused. Properly, S should be understood as the generator of rotations (the actual element of the SU(2) algebra) while \sigma denotes the Pauli matrices. The relationship between them (for spin 1/2 only) is S=\sigma/2 for each component x, y, and z. With this in mind, the ... 2 I have asked you to search phys.SE for the "rigged Hilbert space" words. There should be a lot of returns, the most recent which addresses your second point is the one linked here on this page (right side): Conceptual question on eigenvectors in quantum mechanics. It is pretty good, apart from a technical detail that is not so important to you. As ... 2 Question asked in the title and quoted in the post are different. For the one in the title you have: A\psi=\lambda \psi and [A,B]\psi=0 and asking if \psi is an eigenfunction of B - [A,B]\psi=0\Rightarrow AB\psi=BA\psi=B\lambda\psi=\lambda B\psi \Rightarrow B\psi  is an eigenfunction of A with eigenvalue \lambda. If A is non-degenerate in \... 2 The states \psi_n are not eigenstates of the operator \hat{Q}, only the Hamiltonian. Thus, while \psi_n satisfy$$\hat{H}\psi_n = E_n \psi_n,$$in general$$\hat{Q}\psi_n \not \propto \psi_n.$$Your job is to find a bunch of states \phi_n which are eigenstates of the \hat{Q} operator. i.e., a bunch of states \phi_n that satisfy:$$\hat{Q} \phi_n =...

2

Your first formula is wrong. The Lagrange shift operator dictates $$e^{a\partial_x} \psi (x)=\psi(x+a).$$ Now, $$\hat p|p\rangle = p|p\rangle$$ is a definition of $|p\rangle$, so $e^{ia\hat p} |p\rangle = e^{ia p} |p\rangle$. You then have $$\langle x|e^{ia\hat p} |\psi\rangle = \langle x|e^{ia\hat p}\left ( \int dp |p\rangle \langle p|\right ) |\psi\... 2 Inside the integrals, everything is a scalar, you can rearrange terms as you wish. It's a bit hard to see because you omitted the x-dependence of \psi. It really is$$\int \psi^*(x)\,x\,\psi(x) dx$$where \psi(x) = \langle x|\psi\rangle, which clearly is a complex scalar variable, so swap stuff around any way you like. 2 Taking Hermitian Conjugate is simply taking Complex Conjugate and then Transpose, both operations are linear so their composition is also linear:$$(\hat{A}+\hat{B} )^{\dagger}=((\hat{A}+\hat{B} )^{T})^{*}=(\hat{A}^{T}+\hat{B}^{T} )^{*}=(\hat{A}^{T})^{*}+(\hat{B}^{T} )^{*}=\hat{A}^{\dagger}+\hat{B}^{\dagger}=\hat{A}+\hat{B}  In ...

1

$\psi$ has a definite value for B if and only if $B|\psi\rangle = b|\psi\rangle$. That means: $AB|\psi\rangle = A (b |\psi \rangle) = bA|\psi\rangle$ Using the same logic, of course, $A|\psi\rangle = a|\psi\rangle$, so $bA|\psi\rangle = ba|\psi\rangle$ From here you can see that $BA|\psi\rangle = ba|\psi\rangle$, making the commutator of A and B clearly 0. ...

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