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 ...


7

It's good that you're confused because Susskind's notation is ridiculous. $\psi(x)$ is a number and so you cannot conceivable apply the $\hat x$ operator to it. This is an example of typical misuse of notation by physicists who like to denote a function $f$ by its value at a particular point $f(x)$. This abuse of notation is responsible for so much confusion ...


5

Actually it doesn't have to be from a mathematical perspective, but this only comes up in $3$ or more particles. Let's consider the case of $3$ particles as it's enough to illustrate the point. For permutation group $S_3$ there is a 2-dimensional representation, meaning there are states $\psi(x_1,x_2,x_3)$ and $\phi(x_1,x_2,x_3)$ for which $P_{ij}\psi(x_1,...


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 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

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

It’s not $\nabla$ which is behaving strangely, it’s $\frac{d}{d x}$. You note that $A\cdot\nabla\ne\nabla\cdot A$, by which you mean $A\cdot\nabla f\ne\nabla\cdot(A f)$ in general for a test function $f$. But it’s also true that $g\frac{d}{d x}\ne\frac{d}{d x}g$, in the same sense that $g\frac{d f}{d x}\ne\frac{d}{d x}(g f)$ in general. That second ...


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

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


4

That isn't what happened in line 4 to 5. The px in line 4 is replaced by xp + [p,x] (because px - xp = [p,x]) and results in the additional term to the right of the first term. This is a very common gambit in dealing with commutators. It is the only "legal" way to switch the order of non-commuting operators. Like this: $$ \begin{equation} line 4....


4

My question therefore is why is it appropriate to use the $|+\rangle$ and $|-\rangle$ states when they are the eigenstates of the $H$ operator and not the $U(t)$ one? The eigenvectors $|+\rangle$ and $|-\rangle$ of $\hat{H}$ are also eigenvectors of $\hat{U}(t)$, because $$\begin{align} \hat{U}(t) |+\rangle &=e^{it\hat{H}/\hbar} |+\rangle \\ &=\sum_{...


4

What is the difference between the operator $\hat{p}^2$, and the operator $\widehat{p^2}$? For the sake of clarity let me call your operator $\widehat{P^2}$ as $\hat{M}$ (and let us first work with one dimension). Then by your definition (as stated in the comments), $$\hat{M}\vert\psi\rangle = p^2 \vert\psi\rangle,$$ where $\vert\psi\rangle$ is a momentum ...


3

My question is, what is so special about the eigenstates of this operator and why do they correspond to particles? In QM when identical particles are indistinguishable we require the quantum state describing the particles to remain the same under particle exchange. Since multiplying a state vector by a constant doesn't change the actual state, this is ...


3

Hints: Note that there are implicitly written radial order ${\cal R}$ and normal order $::$ at various places in OP's equations. The starting point is the 2-point relation $$\begin{align}{\cal R}(A(z)B(w)) ~-~:A(z)B(w): ~=~& C(z,w)~{\bf 1}, \cr C(z,w)~\equiv~&\langle \Omega | {\cal R}(A(z)B(w))|\Omega\rangle,\end{align} \tag{1} $$ cf. this Phys.SE ...


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

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

The key results to remember are \begin{align} \langle x\vert p\rangle &= \frac{1}{\sqrt{2\pi\hbar}}e^{i p x/\hbar}\, \\ \hat p\langle x\vert p\rangle &=-i\hbar\frac{d}{dx}\langle x\vert p\rangle \, ,\\ \psi(x)&=\langle x\vert\psi\rangle\, ,\\ \psi(p)&=\langle p\vert\psi\rangle. \end{align} Thus, \begin{align} \hat x \langle p\vert x\rangle : =...


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

Note that he drop the subscript $\langle\rangle_{S^{2}}\rightarrow\langle\rangle$ when he write $\langle (v+q)(\tilde v+\tilde q)\rangle$ so you should not interpret this as a true correlation function. It is just a notation tool that means: sum over all contractions of $q's$ using $-\eta^{\mu\nu}(z-z')^{-2}\alpha '/2$ so $$ \langle \partial x^{\mu}(z) \...


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'm not sure I completely understood your question, but perhaps it will become clear if I explain what the last equation means. The left hand side of the equation means: Take a state $|\mathbf{x}'\rangle$ Translate it by some infinitesimal amount $\text{d}\mathbf{x}'$ using the translation operator Act on it with the position operator, $\mathbf{\hat{x}}$. ...


2

Actually in 2a there should be \begin{align} \langle n|a^\dagger a +a^\dagger a^\dagger a a|n\rangle & =n+\langle n|a^\dagger a^\dagger \sqrt{n-1}\sqrt{n}|n-2\rangle \\ & =\sqrt{n}\sqrt{n-1}\sqrt{n-1}\sqrt{n} \\ & =n+n(n-1) \end{align} I had a feeling there was something more complicated about the powers of number operator but fortunately not.


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

Probably it is cleaner to do it by series. \begin{equation} \begin{split} U^\dagger(t)&=\left(\sum_{n=0}^\infty \frac{1}{n!}\left(\frac{-it}{\hbar} \right)^n H^n\right)^\dagger\\ &=\sum_{n=0}^\infty \frac{1}{n!}\left(\left(\frac{-it}{\hbar} \right)^n \right)^\dagger (H^n)^\dagger\\ &=\sum_{n=0}^\infty \frac{1}{n!}\left(\frac{it}{\hbar} \right)^n ...


2

This process is valid, but indirectly so. Like, if you were my student and you handed it to me I would be very worried that you had a “magical” attitude towards the mathematics involved, whereas this is something more than just an “umbral” similarity between two different mathematical domains. What is actually happening is that you have a time-independent ...


2

They're all the same. $$\vec a \cdot \nabla \vec b \equiv (\vec a \cdot \nabla) \vec b \equiv \vec a \cdot (\nabla \vec b)$$ In cartesian coordinates, the $\nabla$ operator is defined as: $$ \nabla = \left( \begin{array}{ccc}\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \end{array}\right) $$ When this ...


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

The Schroedinger equation is an equation in terms of state vectors in a Hilbert space and reads: $$ i\partial_t |\psi\rangle = \hat{H} |\psi\rangle$$ For some hermitian operator $\hat{H}:\mathcal{H}\to\mathcal{H}$. It is usually possible to split a Hamiltonian into a 'kinetic' part and a 'potential' part: $$\hat{H}=\hat{T}+\hat{V}$$ These are both operators ...


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