15

We mean the second option. For the same eigenvalue $a_i$, there are multiple linearly independent eigenvectors $\psi_{ij}$ where $j$ denotes the degeneracy. When you measure something, if there are multiple linearly independent states giving the same measurement value, those states are degenerate. Also note that the same eigenstate cannot have multiple ...


5

We mean (2) : for the same eigenvalue $a_i$ there are more than one eigenfunction $\psi_i$. In this case one needs an additional index to distinguish different eigenfunctions corresponding to the same eigenvalue: $$\hat{A}\psi_{n\nu} = a_n\psi_{n\nu}.$$


5

If $[\hat A,\hat B]=0$ and they are both non-degenerate, then every eigenstate of $\hat A$ is an eigenstate of $\hat B$ and vice versa. If $[\hat A,\hat B]=0$ and $\hat A$ has a degenerate spectrum, then you are guaranteed the existence of one common eigenbasis. However, you are not guaranteed that every eigenstate of $\hat A$ will be an eigenstate of $\hat ...


5

We are in principle trying to solve the angular TISE problem$^1$ $$ \vec{\bf L}^2Y~=~\hbar^2\ell(\ell+1)Y, \qquad {\bf L}_zY~=~\hbar m Y, $$ on the unit 2-sphere $\mathbb{S}^2$. However, we are using a "tropical" coordinate system $(\theta,\phi)$ that is singular at the north & south poles $\theta=0,\pi$. Hence, we should strictly speaking also ...


5

Actually, unless dealing with Weyl operators, dealing with bosonic fields on a given spacetime $M$, the field operators are elements of a $*-$ algebra with unit $\cal A$. Abstract boson field operators are algebra-valued linear functions $$C_0^\infty(M) \ni f \mapsto \phi(f) \in \cal A \:,$$ satisfying some further properties. The algebra is made of finite ...


4

You are correct $-$ the book is (slightly) abusing notation here in the name of simplicity. As you point out, the term 'operator' is generally understood to be a linear mapping $\hat O:\mathcal H \to \mathcal H$ from one vector (Hilbert) space to itself. The gradient, on the other hand, is what's known as a vector operator, which is basically a triplet of ...


4

The operator $\hat\sigma_{11}$ is a projection operator onto the space spanned by $|1\rangle$. Geometrically it's easy to see that the projection of a vector will always be shorter than the vector itself, and this is true in general for projection operators. So the expectation value $\langle\hat\sigma_{11}\rangle$ is necessarily smaller than one. Hence $\...


4

Your mistake is an abuse of notation where you say $$\delta (ax - ax^\prime) = \langle ax \vert ax^\prime \rangle = a^* a \langle x \vert x^\prime \rangle$$ In the first equality, you are treating $\lvert ax \rangle$ as the eigenstate with eigenvalue $ax$, but in the second equality, you are treating it as $a\lvert x \rangle$, the eigenstate with eigenvalue $...


4

We should have the operator equation $\partial_\mu \hat J^\mu=0$. This implies that $\langle a|\partial_\mu \hat J^\mu|b\rangle=0$, $\forall a,b$.


3

In QM lectures happens often that a certain operator is described as acting on ket vectors $A|\psi\rangle$ and then after a bit the same operator, without any further explanation, is shown as acting on functions $A\psi(x)$. This is not correct. You may have seen it somewhere, but the author was being sloppy or abusing notation. Let $|\psi\rangle$ be an ...


3

I believe the issue here is with a loose use of term operator. Gradient is a mathematical operator, i.e. it is an operator in the same sense as arithmetic operations, divergence operator, curl, integration, etc. - it transforms a mathematical function. It is however not an oprator in the sense of linear algebra (a definition given in the question), which is ...


3

From a more mathematical point of view, we say there is degeneracy when the eigenspace corresponding to a given eigenvalue is bigger than one-dimensional. Suppose we have the eigenvalue equation $$ \hat A\psi_n = a_n\psi_n\;. $$ Here $a_n$ is the eigenvalue, and $\psi_n$ is the eigenfunction corresponding to this eigenvalue. But this eigenfunction is of ...


3

In your case you seem to have defined $\phi_i = \hat{B}\psi_i$, where $i=1,2,3,\dots N$ is the degree of degeneracy. It should be clear to you that the states $\phi_i$ are still eigenstates of $\hat{A}$. However, there is no reason for them to, a priori, be eigenstates of $\hat{B}$. In fact, since every $\phi_i$ is an eigenstate of $\hat{A}$, you can write ...


3

The quantum version of the Noether theorem is Ward-Takanashi identity - https://en.wikipedia.org/wiki/Ward%E2%80%93Takahashi_identity. It gives relationships between correlation functions in the QFT : $$ \langle \partial^{\mu} j_{\mu} (x) \mathcal{O}_1 (x_1) \ldots \mathcal{O}_N (x_N) \rangle = \sum_{n = 1}^{N} \delta(x - x_n)\langle \mathcal{O}_1 (x_1) \...


3

I do not know how it is implemented in a Josephson junction, but if we were working with photons, one way to realise the action of the Hadamard gate is by using a perfect beam splitter. Operator In particular, the Hadamard gate is given by: $$H=\frac{1}{\sqrt{2}}\left[\begin{array}{cc} 1 & 1 \\ 1 & -1 \end{array}\right]$$ In comparison, the action of ...


3

The result of an inner product is a scalar--that is, a 1x1 matrix. Transposing a scalar leaves it unchanged. So, $(B^{*T}A)^{*T} = (B^{*T}A)^{*}$. On another note, the usual symbol for the complex transpose is the dagger $^{\dagger}$, \dagger in LaTeX.


3

In David Tong: Lectures on Quantum Field Theory, section 6.2 he discussed 2 ways of Maxwell theory quantisation in different gauge: Coulomb Gauge Lorenz gauge. The Lorenz gauge is a way of Lorentz invariant quantisation, so he gave more details on such a quantisation. But he gave all sufficient information about Coulomb Gauge quantisation. The essential ...


2

When one of the two commuting operators has degenerate eigenfunctions, one can always construct their linear combinations which will be the eigenfunctions of the other operator.


2

A function $\psi$ is called bounded on some region $R$ if there exists some $C>0$ such that $|\psi(x)|<C$ for all $x$ in $R$. This does not imply that the function is normalizable; for example, the function $\psi(x)=1$ is bounded on the entire real line, but is clearly not normalizable. This is also different from the author's definition of a bound ...


2

You can read about canonical experiments like: The Hong-Ou-Mandel experiment. There are also YouTube videos like this one or this one but just searching for HOM experiment in Google will return multiple hits. The possible outcome of a scattering experiment is dictated by the amount of superposition of wavepackets inside an inteferometer. This is also a &...


2

The standard hypotheses for the valuation function $v: B(H)_{sa} \to \mathbb{R}$, where $B(H)_{sa}$ denotes the real linear subspace of the selfadjoint operators of $B(H)$ and $2< \dim(H) < +\infty$, are that (i) $v(A+B) = v(A) +v(B)$ for all $A,B\in B(H)_{sa}$ with $AB=BA$; (ii) $v(AB) = v(A)v(B)$ for all $A,B\in B(H)_{sa}$ with $AB=BA$. The thesis ...


2

As a quick notational point, given some vector space $V$ over a field $\mathbb F$, $V^*$ typically refers to the algebraic dual space, i.e. the vector space of linear maps from $V\rightarrow \mathbb F$. If $V$ is in particular a topological vector space (so it has some notion of continuity, convergence, etc), then we can define the topological dual space $V'...


2

The two parts of your question illustrate two of the motivations for the AQFT approach: to focus on observables instead of on fields, and to use only bounded operators — ordinary operators that are well-defined on the whole Hilbert space. In a canonical formulation of QFT, field operators are the fundamental objects (mathematically), and observables are ...


2

You have had this question answered multiple times in numerous questions you have asked in the last couple of weeks on the Dirac notation, and operator action on bras (not kets, as promised!), $$\langle x | \hat{p} = -i\hbar \partial_x \langle x | , \tag{3}$$ whence $$\langle x | \hat{p}^2 = (-i\hbar \partial_x)^2 \langle x | , \tag{4}$$ and therefore $$...


2

The weaker notion will also allow for anti-unitary operators, such as time-reversal and charge conjugation: https://en.wikipedia.org/wiki/T-symmetry#Time_reversal_in_quantum_mechanics These transform the amplitude to its complex conjugate, keeping the norm the same. They can be very interesting, such as in the CPT theorem in relativistic QFT.


2

It is simply non-physical. Wigner symmetries deal with rays $[\psi]$, i.e., non-vanishing vectors of the Hilbert space up to multiplicative factors. $$[\psi] := \{a\psi \:|\: a \in \mathbb{C} \setminus \{0\}\}\:, \quad \psi \in H \setminus \{0\}$$ Transition probabilities of couples of rays are defined as $$P([\psi], [\psi']) := \frac{|\langle\psi|\psi'\...


2

Assuming the $\mid k \rangle $ are orthogonal, you have written down a unitary operator. A matrices is an examples of normal operators and normal operators have a spectral decomposition. It is not so standard, but one can consider unitary or even normal operators as observables, but they are complex-values observables. More standard is to use two real ...


2

Maybe I could help you with your confusion. If we consider two hermitian operators (just ordinary QM) $A,B$ and regard $$<v|AB|v>$$ you could run into several scenarios. $|v>$ is eigenstate for operator $A$ (with eigenvalue $a$) and operator $B$ (with eigenvalue $b$). Then your argumentation is correct. $<v|AB|v>=ab<v|v>=<v|A|v>&...


2

It is generally NOT true that $B|x\rangle=b|x\rangle$. This is only true if $|x\rangle$ is an eigenvector of $B$. However, there is another flaw in your reasoning. Even if $|x\rangle$ is an eigenvector of both $A$ and $B$ (as you assumed), it is NOT true that $\langle x| x\rangle=1$ (I assume $|x\rangle$ represents an eigenvector of the position operator $X$)...


2

$\langle \cdot | \cdot \rangle$ is an inner product. First let's rename $r$ as $r_0$ and lets think of it as a fixed constant (i.e. one meter or whatever you want). When you write $\langle r_0 | \psi \rangle$, you are taking the inner product between your state $| \psi \rangle$ and $\langle r_0 |$. But $| r_0 \rangle$ is an element of the basis $\{| r \...


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