Tag Info

New answers tagged

1

You are maybe making confusion between the action of an observable (operator), and the measurement process. In particular: $A\psi$ is simply a vector of the Hilbert space. In my opinion it has not much sense of talking about "initial" and "terminal" state because you are not looking at a dynamical situation. If you want to know the average value of an ...


1

From the equations, $\phi$ is the operator acting on the variable/state $\xi$. It is important to notice also that in the factorization the $a_i$ numbers are real and the $c_i$ are complex. This factorization comes just from the mathematical fact that for a polynomial equation of degree $n$, such as the first equation, there are $n$ complex roots.


1

No $\hat\phi|0\rangle$ is not an eigenvector of $\hat\phi$. You can see this, for example, by writing out $\hat\phi$ in terms of creation and annihilation operators, then compare $\hat\phi|0\rangle$ against $\hat\phi^2|0\rangle$, and observe that one is not a scalar multiple of the other. So as you suspected, eq. 5 is not correct To obtain some analogy of ...


3

You can utilize the construction of the $Q$ space, as described in Reed and Simon vol.2, page 228-230. Oversimplifying, you can make the analogy $\lvert \phi\rangle \sim \lvert x\rangle$, but the associated momentum is not $\hat{p}$, but $\hat{\pi}$ (the canonical conjugate momentum of the field $\hat{\phi}$). With slightly more precision: the Fock space ...


3

$$D(\hat{X}_i):= \left\{ \psi \in L^2(K, d^nx)\:\left|\: \int_K x_i^2|\psi(x)|^2 d^nx< \right.+\infty \right\} = L^2(K, d^nx)$$ where the last identity holds true if $K$ is bounded (in particular compact) because $x_i^2$ is bounded as well thereon (in this case the operator is bounded, too). Moreover $$D(\hat{J}^2):= \left\{\psi \in L^2(\mathbb S^2, ...


0

To answer the question I think you meant to ask, $V(x,t)$ is a real-valued function. But you can also think of it as an operator. An operator is anything that maps functions to other functions, and multiplication by a fixed function is one way to do that. To be a little more pedantic with the notation, if you use $\hat{V}$ to represent the operator, then ...


0

As usually, physicists are being a bit sloopy with notation here. Mathematically, there's a big difference between a function (such as $V$) and a function's value for some arguments (such as $V(x,t)$). The latter is indeed simply a number, scalar if you like. However, in physics, functions are often used just as "families of values" and values with some free ...


3

In the representation given by you, the $V(x,t)$ is a scalar function depending on $x$ and $t$. However, you already have choses a basis (the $x$-basis). In general, the $\hat{V}$ or rather $\hat{H}$ in the Schödinger Equation is an operator. This operator can then be evaluated in a basis of your choice. If you are familliar to the dirac notation one can ...


3

The potential is, without any ambiguity, an operator. There is no other way, because it changes the spatial dependence of the wavefunction: $$ \Psi(x,t)\text{ changes to }V(x,t)\Psi(x,t). $$ It's unclear to me what you mean by "scalar", though, because this word can have multiple meanings. For example, it is indeed a scalar operator - as opposed to vector ...


0

Answer to question one : The Principle of Quantum Mechanics by R. Shankar page 149 reads "Barring a few exceptions, the schrodinger equation is always solved in a particular basis. Although all basis are equal mathematically, some are more equal that others. First of all, since H = H(X,P) the X and P basis recommend themselves....The choice between the two ...


0

You have to clarify the term "basis" in infinite dimensions. By definition, finite-dimensional spaces have a finite basis (so that, for instance, $\mathbb{R}^n\approx\mathbb{R}^m$ iff $n=m$). But for infinite-dimensional spaces, there are various types of bases (assuming you can take infinite sums). A natural one (using finite sums) is the Hamel basis, but ...


5

First, the term "basis spaces" isn't standard in quantum physics but let us assume that we understand what the sentences approximately mean. Second, the momentum is continuous (not quantized) if the position space is noncompact (infinite). The momentum only becomes quantized if the position space is compact (or periodic), and indeed, it's been ...


2

An observable is a self-adjoint operator $\mathcal{O}$ on the Hilbert space of states $\mathcal{H}$. The spectral theorem tells us that such an operator has an orthonormal basis of eigenvectors in $\mathcal{H}$, if it is compact. If it is not compact, we have to "enlarge" the Hilbert space to something called rigged Hilbert space or Gelfand tripel. A ...


0

Unfortunately not every operator, which is used to get an observable gives you enough elements to make a base of the Hilbert space. As far as I know every eigenstate is orthogonal to each other. But the “length” could be arbitrary. Mostly used eigenstates in quantum mechanics are normalized for better usage.


3

The Projection Operator $\delta_{\varepsilon\, a}$ is $1$ for some particular $|a\rangle$ and is $0$ on any orthogonal state. The definition of the expectation in classical probability theory for this operator is given by \begin{equation} \left\langle \delta_{\varepsilon\, a}\right\rangle = 1\Pr(|x\rangle = |a\rangle) + 0\Pr(|x\rangle \perp |a\rangle) = ...


4

The position operator in three dimensions is a vector operator, which, as in mpv's answer, acts as $$ \hat{\mathbf r}\psi(\mathbf r)=\mathbf r \psi(\mathbf r). $$ Note that the hat denotes an operator and not a unit vector. The definition of a vector operator is somewhat tricky, and it is indeed startling that an operator can have vector eigenvalues. The ...


3

The position operator in 3D is a vector in 3D: $$ \hat{\bf r} \psi({\bf r}) = {\bf r} \psi({\bf r}) $$ See here.


2

The first equation is quite elementary to derive. First, we define the expectation value of an operator: $$\langle O \rangle=\langle \psi(t)| O | \psi (t)\rangle = \langle \psi(t=0)|U^\dagger O \ U |\psi (t=0)\rangle \tag{1}$$ where $U$ is the time evolution operator: $U=\exp\left( -i\frac{Ht}{\hbar}\right)$ if $H$ is independent of time. Now, we can take ...


2

The second one is a particular case of the first one, as I understand it. In the first one we are asuming that, in general, Q can depend on time explicitily and through other dynamical variables, as for example, position or momentum. That is, $$ \hat{Q} = \hat{Q}(\hat{x}(t),\hat{p}(t),t) $$ So, if we want to know how it evolves with time, we must to know ...


0

For your second problem, the propagator can be written with its indices as $$ (S_F)_{\alpha\beta}(x-y) = \langle T\{\psi_{\alpha}(x)\bar\psi_{\beta}(y)\} \rangle $$ Then we have $$ \langle T\{\bar\psi(x)\Gamma\psi(y)\} \rangle = \Gamma_{\alpha\beta} \langle T\{\bar\psi_{\alpha}(x)\psi_{\beta}(y)\} \rangle = -\Gamma_{\alpha\beta} ...


6

Dirac being opaque and hard to follow? Well I never... In Chapter 10 Dirac argues on physical grounds that the eigenkets of an observable must form a complete set. His argument goes that say we have an observable with eigenstates $|\varepsilon\rangle$ and some general state $|P\rangle$. Then I can write $|P\rangle$ as \begin{equation}|P\rangle = \sum ...


0

I) Before we start let us briefly recall certain aspects of the formalism from Ref. 1. The Minkowski sign convention is $(+,-,-,-)$. The 'phase space' measure for a particle $A$ is $$ d\tilde{k}~:=~ \frac{d^3k}{(2\pi)^3 2\omega_{k,A}} ~=~ \frac{d^4k}{(2\pi)^3} \delta(k^2-m^2_A)\theta(k^0), $$ $$ \omega_{k,A}~:=~\sqrt{{\bf k}^2+m^2_A}~>0~. \tag{3-35}$$ ...


0

You obtain this by Wick's Theorem, which can be stated as $$T\{\phi_1\phi_2...\phi_n\}=N\{\phi_1\phi_2...\phi_n+\sum\text{all possible contractions of }\phi_1\phi_2...\phi_n\}$$ where N is the normal ordering operator which puts all the daggered fields on the left ( for example ...


1

Recall the Heisenberg principle which tells us that a we cannot know both position and momentum at the same time $$\Delta x \Delta p \geq \hbar/2$$ For a particle, we have actually at most $\Delta x = L$, so that we cannot in principle have a sharp momentum state. Fortunately, there are eigenstates which are eigenstates of $\hat{P}^2$, that is, states ...


1

Taking the standard $[0,L]$ problem, eigenfunction and energy eigenvalues are: $$ \varphi_n=\sqrt{\frac{2}{L}}\sin\frac{n\pi x}{L}, \qquad E_n=\frac{\hbar^2\pi^2n^2}{2mL^2}. $$ This means that stationary 1D box systems (e.g. insulated ones) only admit states with a discrete set of possible energies, as above. Now, that as far energy is concerned. What about ...


0

Your lecture notes, or your transcription of them, are in error. You should have $$ \hat{L_{+}}|\lambda,m_\mathrm{max}\rangle= 0 \\ \hat{L_{-}}|\lambda,m_\mathrm{min}\rangle= 0 $$ That is, raising the maximum-projected state doesn't give you $\left|\lambda,m\right> = \left|0,0\right>$, a state with no angular momentum, and it doesn't give you a vacuum ...


0

Sometimes it's helpful to consider finite dimensional special case to get insights on a problems like these. Let $P$ and $L$ be in $\mathbb{C}^{n \times n}$. If $P^2 = P$, then $Pv = \lambda v$ leads to $\lambda^2 = \lambda$, i.e., the eigenvalues are in $\{0, 1\}$. In the context of quantum, we consider Hermitian operators. In this case, we can find some ...


0

I think you need to add a postulate to (3) in order to obtain (1). This postulate it's actually assumed in every picture of the dinamic and it is that the evolution operator is a one group parameter. It is such a natural assumption that it comes no harm in taking it for true. Even in classical hamiltonian mechanics the flow in the coordinate space form a ...


4

The result can be proved in a general way using, well, math. In particular the theory of semigroups of linear operators on Banach spaces (I know that seems advanced and maybe not physical, but it is an elegant way of proving the result you seek ;-) ). Define the Banach space $\mathscr{L}^1(\mathscr{H})$ of trace class operators over a separable Hilbert ...


1

Just calculate (3) for $\hat{A}_H=\hat{U}_H$. With $\hat{U}_H(t)=\hat{U}^{\dagger}(t)\hat{U}(t)\hat{U}(t)=\hat{U}(t)$ and $\hat{H}_H=\hat{H}$ (as you said) you get: $\frac{d\hat{U}_H}{dt}=\frac{\partial \hat{U}}{\partial t}+\frac{1}{i\hbar}\underbrace{[\hat{H},\hat{U}]}_{=0 \text{(as you ...


0

Dirac already considered it in a 1925 paper(Open access). (Thanks to Alex Nelson for the comment to the OP's question. I found it at the Wikipedia's article on Groenewold. =]) Operator ordering issue also exists in path-integral formalism, and of course Feynman was aware of it. See for example "Techniques and applications of path integration" by Schulman. ...


4

CAUTION - ANSWER INCOMPLETE There is a gap in my argument (see the send); it relies on the claim that \begin{align} - \hat O^\dagger \hat A = \hat A\hat O \end{align} for all hermitian $\hat A$ implies $\hat O = 0$ which may not be true. Please comment if you know how to prove this or know of a counterexample. Update. Actually the claim above is ...


1

Without further constraints you cannot say anything about $f$. Observe that the identity is idempotent and $id \circ f = f$ holds for any function $f$ on any set. Alright. If $P$ is not the identity, it cannot be invertible, thus $\ker(P) \neq 0$. Idempotent operators are basically projection operators, they are diagonalizable and only have eigenvalues $0$ ...


2

The energy operator is obtained via the so-called correspondence principle. This means that one considers the classical expression for the total energy $$\frac{p^2}{2m}+V(x)$$ and replaces the momentum and position variables (numbers in classical mechanics) by the momentum and position operators. $p^2/2m$ is the kinetic energy (it's just another way of ...


2

The "Energy operator" in a quantum theory obtained by canonical quantization is the Hamiltonian $H = \frac{p^2}{2m} + V(x)$ (with $V(x)$ some potential given by the concrete physical situation) of the classical theory promoted to an operator on the space of states. Since the core of the quantization procedure is promoting the classical phase space ...


1

By using an arbitrary test function you haven't included any information about the state of the electron. This means you are considering all possible states of an electron in a coulomb potential, so that formula you derived is true for all bound states of the hydrogen atom and all the unbound states of an electron scattering off the proton. It is true in the ...


0

A tricky question, really. Apart from the fact that your $\lvert e,p\rangle$ vector does not belong to $L^2$ (hence you cannot take scalar products of it), I don't see any other flaw. That, in my opinion, means you have a nice argument to prove the following mathematical statement: Let $\mathscr{H}$ be a separable Hilbert space, $0\neq z\in\mathbb{C}$. ...


3

It's not non-sensical at all, except that there shouldn't be a minus sign (as mentioned in the comments) and that you took an operator outside of an expectation value, which I think worked out OK in this case but in general should be avoided. More conservatively, $$ \hat x = i \hbar \frac{d}{d \hat p} $$ it follows that $$ \langle q \mid \hat x \hat p ...


8

As is customary in such question, I will point to this paper, which excellently discusses the problems the Dirac formalism has. Now, in your concrete example, the problem lies in the energy/momentum states $| p_0 \rangle$ themselves, which are non-normalizable, since the wave function associated is the Fourier transform of $\delta(p-p_0)$, which means that ...



Top 50 recent answers are included