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1

The conservative answer is: It depends on context. Different authors mean different things. See also this Phys.SE post.


3

It is definitely an ambiguous notation, but one that is quite conventional. You should interpret it as: $(\partial_a X_\mu)(\partial^a X^\mu)$. For instance, often the Klein-Gordon Lagrangian is written as: $$\mathcal{L} = \frac{1}{2} \partial_\mu \phi \partial^\mu \phi + \cdots $$ which should be interpreted as: $$\mathcal{L} = \frac{1}{2} (\partial_\mu ...


1

In my experience, the correct interpretation is $$ \partial_\mu X \partial^\mu X = (\partial_\mu X)( \partial^\mu X). $$ Total derivatives are usually written clearly as $$ \partial_\mu(\ldots). $$


2

We interpret OP's question (v4) as: How do we recover the phase ambiguity from the generator of translation method in Ref. 1? Recall that an eigenvector for an operator can be rescaled with a non-zero multiplicative factor. The main point is that the position eigenket $| x \rangle$, which satisfies $$\tag{A} \hat{x}| x \rangle~=~ x| x \rangle, $$ ...


2

Usually, when you want to probe the lifetime $\tau$ of a particle (or a quasi-particle), what you basically do is looking for the way the associated wave-function $\psi$ is significantly decreasing : $$\psi\sim\psi_0\,e^{-\left(\frac{1}{\tau}+\,i\frac{E}{\hbar}\right)t}$$ Let us consider a free particle with a given energy $E$. As $\langle E\rangle=E$ is ...


0

In Mahan may-particle physics P15: (for bilinear Hamiltonian)It is only necessary to find the eigenvalues of the Hamiltonian matrix. Usually the matrix is of infinite dimensionality. But one may often diagonalize it exactly for many problems. Computers allow very accurate solutions for any case of interest. If all Hamiltonians had only bilinear ...


0

"projection operators commute → they're the same" Are you sure he said this predicate ? or it is your own consequence? However, it is not true ! Consider two dimensional X-Projector And Y-Projector , they commute but they are not the same!


2

Suppose we have arbitrary vectors $|\alpha\rangle$ and $|\beta\rangle$ that are not necessarily aligned with one another. We can determine the component of $|\beta\rangle$ that lies along the direction of $|\alpha\rangle$ by defining an operator $$ \hat{P}=|\alpha\rangle\langle\alpha| $$ which we call the projection operator. Note that $\hat{P}$ is ...


0

A complete set of eigenstates spans the whole space, not just the subspace the projection operators project on. In this set of eigenstates you also have a basis of the subspace belonging to the eigenvalue 0.


3

Seeing as $$\langle k|k_1k_2\rangle = \langle 0| a(\mathbf{k}) a^{\dagger}(\mathbf{k_1}) a^{\dagger}(\mathbf{k_2}) |0\rangle$$ and $$a(\mathbf{k})a^{\dagger}(\mathbf{k_1}) = a^{\dagger}(\mathbf{k_1})a(\mathbf{k}) + f(\omega)\delta(\mathbf{k}-\mathbf{k_1})$$ you'll get one term that vanishes because you cannot destroy the vacuum and one term that simply ...


4

If you are interested in physical applications you could also include: Bratteli-Robinson: Operator algebras and quantum statistical mechanics It is a two-volume quite complete book, mathematically minded, discussing lots of applications of operator algebras theory to several physical systems, especially arising from statistical mechanics. Haag: Local ...


1

We know how the creation operator acts on the vacuum, i.e. $$(a^{\dagger})^n \lvert0\rangle =\sqrt{n!}\lvert n \rangle$$ where we use the notation $\lvert n \rangle$ to signify the state such that $\langle x \lvert n \rangle = \psi_n(x)$, i.e. the $n$th energy level of the harmonic oscillator. The annihilation operator kills the vacuum. Therefore the ...


2

As you said if two operators commute they share eigenvectors. Physically this means that you can have a definite value for both. For example in the hydrogen atom the Hamiltonian $H$, which is the energy, and $J^2$, the magnitude of angular momentum, commute. A hydrogen atom can be in a state of definite energy and definite angular momentum. However, the ...


0

There are two things I don't understand in this derivation. The first is that the total time derivative $\frac{d}{d t}$ turns into a partial time derivative $\frac{\partial}{\partial t}$ when inside the integral. Generally speaking, this is implied by the Leibniz intregral rule, which is a particular case of differentiation under the integral sign. For ...


1

Neither left-hand side nor right-hand side are properly operators that act on states and give states. They are a form of operators that act on states and give complex numbers. We call such a thing a bra (and a state is called a ket, so when you pair them you get a bra(c)ket). The left-hand side acts like $$((H\psi)^*)(\varphi) = \int \psi^* H \varphi \, ...


0

In finite dimensions, $$\langle\psi_n|U^\dagger U|\psi_m\rangle$$ extracts the $(n,m)$ component of $U^\dagger U$ in basis $\{|\psi_n\rangle\}$. Since the result is $\delta_{nm}$, so $U^\dagger U = I$. In infinite dimensions, you can apply double-integrals $$\int_n \int_m |\psi_n\rangle(\cdot)\langle\psi_m|\; dm\; dn = I$$ on both sides, separate the ...


0

Starting with the first two identities you can actually show that the last expression is satisfied not just for the basis vectors but for any two vectors that are elements of the space spanned by the basis vectors, i.e. $\langle\eta|U^{\dagger}U|\theta\rangle = \langle\eta|\theta\rangle$. Therefore $U^{\dagger}U = I$.


2

From the Schroedinger equation, $$i\hbar\frac{\partial\psi}{\partial t}=\hat{H}\psi$$ simply divide by $i\hbar$: $$\frac{\partial\psi}{\partial t}=\frac{1}{i\hbar}\hat{H}\psi=-\frac{i}{\hbar}\hat{H}\psi\\$$ where we used $$ \frac{1}{i}\cdot\frac{i}{i}=\frac{i}{i^2}=\frac{i}{-1}=-i $$ Now note that you've incorrectly applied the derivative. In quantum ...


1

They have used the Schrödinger equation: $$i\hbar\frac{\partial \psi}{\partial t} = \hat{H} \psi$$ And hence: $$\frac{\partial \psi}{\partial t} = \frac{1}{i\hbar}\hat{H}\psi$$ And since $1/i = -i$: $$\frac{\partial \psi}{\partial t} = -\frac{i}{\hbar}\hat{H}\psi$$ Similarly for $\psi^*$ we take the complex conjugate: $$\frac{\partial \psi^*}{\partial t} = ...


1

To arrive at the 2nd term of the third line, they have simply employed the SE equation, $$i\hbar\frac{\partial \psi}{\partial t} = \hat{H}\psi$$ Re-arranging the above equation gives: $$\frac{\partial \psi}{\partial t} = -\frac{i}{\hbar}\hat{H}\psi$$ Therefore by replacing $\frac{\partial \psi}{\partial t}$ in the second line with ...


5

First, a note about the Hamiltonian and its time derivatives. I think that it is misleading to write that the Hamiltonian $$ H = i\hbar\frac{d}{dt}, $$ although the time-dependent Schrodinger equation is of course $$ H\psi = i\hbar\frac{d}{dt} \psi. $$ To evaluate e.g. $\frac{d}{dt}H$ you should consider $H=H(p, q, t)$, rather than $H = i\hbar\frac{d}{dt}$. ...


0

It is very common to abuse the notation here, so I'll try to clarify a bit. The state of a physical system can be described by an abstract vector $\left|\psi\right\rangle$, which is an element of a Hilbert space. The wavefunction, $\psi(x)$ is the representation of that vector in the position basis, $\psi(x)\equiv\left\langle x | \psi\right \rangle \equiv ...


1

You need to apply the operator first and then evaluate the integral: $⟨P⟩_ψ = i\hbar\int{\psi^*(x)\frac{d\psi(x)}{dx}dx}$


2

You are right that $\langle A\rangle_{\psi}$ is a number and not an operator. However, people often write just a number when they actually mean the identity operator times that number. So in the right hand side, $\langle A\rangle_{\psi}$ should actually be $\langle A\rangle_{\psi} \mathbf{1}_H$, where $\mathbf{1}_H$ is the identity operator on your hilbert ...


1

The notation is short-hand for an expression utilizing the Backer Campbell Haussdorf formula. Let $X$ and $Y$ be operators, then $$e^{x}Ye^{-X} = Y + [X,Y] + \frac{1}{2!}[X,[X,Y]] + \frac{1}{3!}[X,[X,[X,Y]]] + ...$$ I assume $[X,Y]_{(n)}$ refers to the $n$th term in this expansion; it roughly counts how many times the commutators are nested in each other. ...


3

I think this is the Baker-Campbell-Haussdorff formula, and the notation means to iterate the commutator. That is, $$[L, M]_1 = [L, M]$$ And $$[L,M]_{n+1} = [L, [L,M]_{n}].$$


1

This is not standard notation, and one would typically expect any text that uses it to define it at its first occurrence. Since you understandably cannot provide us with a reference, your best bet is hunting for all occurrences of that notation, starting from there and going up through the text, until it explains what it means. Trust me, it will be there.


0

For Hermitian matrices eigenvectors corresponding to different eigenvalues are orthogonal. This guarantees that not only are the eigenvalues real, expectation values are too.


1

In general, derivative couplings lead to momentum-dependencies in scattering amplitudes. This can be seen from the fact that the Fourier transform of a derivative operator corresponds to a multiplication by the relevant momentum. A mass dependence is implicit through by having a momentum, since the momentum of a fermion depends on its mass. In this case, ...


1

Is it therefore, incorrect to talk about "expectation value of an operator"? Yes, because when you write those integrals you're asking for the average of a dynamical variable $A$ whose associated operator is $\hat A$. The point is that you're asking for a number , e.g. the average position of the electron in a gaussian wavepaket. Then, when you want to ...


4

Remember, operators are nothing but maps. Expectation value of an operator is pretty much defined (I guess) in general operator theory. It just turns out that in QM (Hermitian) operators correspond to dynamical variables. In general you can also calculate expectation values of operators like $L_+$ and $a^{\dagger}$ etc., which don't have any dynamical ...


0

Since the vector $\Lambda | \omega_i \rangle $ has the same eigenvalue as $| \omega_i \rangle $, it must be in the same invariant subspace as $| \omega_i \rangle $, which Shankar takes to be one dimensional.


1

When $\lambda_1$ is an eigenvalue of a matrix and $v_1$ and $v_2$ are the components of the corresponding eigenvector, then the following equation holds: $\begin{pmatrix} a-\lambda_1 & b \\ c &d-\lambda_1 \end{pmatrix}\begin{pmatrix} v_1 \\ v_2 \end{pmatrix}=\begin{pmatrix} 0 \\ 0 \end{pmatrix}$ Now when you scale up the eigenvector (say by three) ...


2

Note that he explains above: "Consider first the case where at least one of the operators is nondegenerate, i.e. to a given eigenvalue, there is just one eigenvector, up to a scale." So he uses the assumption that the operator is nondegenerate and the definition of nondegeneracy (or a statement equivalent to the definition of nondegeneracy, if you use a ...


3

$[\vec\sigma\cdot \vec p, \vec \sigma]_i = [\sigma_j p^j, \sigma_i] = [\sigma_j , \sigma_i]p^j = 2i\epsilon_{jik} \sigma^k p^j= 2i\epsilon_{ikj} \sigma^k p^j =2i(\vec{\sigma} \times \vec{p})_i$. So $[\vec\sigma\cdot \vec p, \vec \sigma] = 2i(\vec{\sigma} \times \vec{p}).$


3

No. The expectation value of the square of the momentum operator cannot be negative. The other answers address your particular problem on an integration level, but also notice that this can be easily shown in bra-ket notation. Let $|\psi\rangle$ be any state in the Hilbert space, and let $\hat P$ be the momentum operator, then we have \begin{align} ...


1

Let's set $\hbar$ and $\alpha$ to one. Then $\psi(x)=C\exp(-|x|)$. Let's compute the first derivative of $\psi(x)$ carefully. $d_x \psi(x) = -C \exp(-|x|)\mathrm{sgn}(x)$. Now let's carefully compute the second derivative of $\psi(x)$. \begin{equation} \begin{aligned} d_x^2 \psi(x) &= -C d_x\exp(-|x|)\mathrm{sgn}(x) \\ &= C ...


2

It seems that OP already knows that the variance is a manifestly non-negative quantity, and he is struggling to explain a negative result that he got. Hint: The wave function $\psi(x)=\sqrt{\alpha}e^{-\alpha|x|}$ is not differentiable in $x=0$. The generalized function $$\tag{1} \psi^{\prime\prime}(x)~=~\left(\alpha^{\frac{5}{2}}- ...



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