# Tag Info

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 $H = i\hbar\frac{d}{dt}$. ...

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 ...

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 ...

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} ...

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 ...

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: $$\mathcal{L} = \frac{1}{2} (\partial_\mu ... 2 We interpret OP's question (v4) as: How do we recover the phase ambiguity from the generator of translation method in Ref. 1? Recall that an eigenvector for an operator can be rescaled with a non-zero multiplicative factor. The main point is that the position eigenket | x \rangle, which satisfies$$\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 ... 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| $$which we call the projection operator. Note that \hat{P} is ... 2 Note that he explains above: "Consider first the case where at least one of the operators is nondegenerate, i.e. to a given eigenvalue, there is just one eigenvector, up to a scale." So he uses the assumption that the operator is nondegenerate and the definition of nondegeneracy (or a statement equivalent to the definition of nondegeneracy, if you use a ... 2 It seems that OP already knows that the variance is a manifestly non-negative quantity, and he is struggling to explain a negative result that he got. Hint: The wave function \psi(x)=\sqrt{\alpha}e^{-\alpha|x|} is not differentiable in x=0. The generalized function$$\tag{1} \psi^{\prime\prime}(x)~=~\left(\alpha^{\frac{5}{2}}- ...

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 ...

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 ...

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 ...

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 ...

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 $$((H\psi)^*)(\varphi) = \int \psi^* H \varphi \, ... 1 They have used the Schrödinger equation:$$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 ...

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) ...

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. ...

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.

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 ...

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{aligned} d_x^2 \psi(x) &= -C d_x\exp(-|x|)\mathrm{sgn}(x) \\ &= C ...

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