Tag Info

New answers tagged

6

Quantum mechanical postulates So is the mathematical expression for each individual operator also a postulate that's not listed, or are they derivable from other axioms? The mathematical expression for each individual operator is sort of a postulate, but it should not be listed. The postulates define a (more or less) complete theory in that I can ...


0

I believe that the important thing to remember is the basic difference between quantum and classical description of a system. In classical mechanics you describe it by defining some sort of equation of motion which gives you a position of all the constituents of the system described by function of time but for each of these particles you have one function. ...


2

The eigenvalue is something physicists should be familiar with. For some matrix, $A$, multiplied by some vector $\mathbf x$, we get $$ A\mathbf x=\lambda\mathbf x \tag{1} $$ where $\lambda$ is the eigenvalue, a characteristic of $A$ on $\mathbf x$. An eigenfunction is related to Equation (1). Given an operator (a differential operator in the case of quantum ...


3

Your "imaginary eigenvalues" don't work, because the eigenfunctions are no eigenfunctions. They do not lie in $L^2$, as you seem to be aware of. So, let's deal with the Laplacian itself: $-\Delta=-\frac{d^2}{dx^2}$. What I want to do is, I want to calculate the Fourier transform of this operator, because the Fourier transform diagonalizes $-\Delta$, as we ...


4

First off, if $k =: i\kappa$ is imaginary, the eigenvalue (“energy”) is $-\kappa^2$, i.e. real but negative: $$-\frac{d^2}{dx^2} e^{ikx} = -\frac{d^2}{dx^2} e^{-\kappa x} = -\kappa^2 e^{-\kappa x}.$$ Physically, that is an evanescent wave in one direction, but grows without bound in the other, so if your space is all of $x\in\mathbb R$, it is not a valid ...


0

A Hermitian operator is diagonalizable and has real eigenvalues. If you have an operator with real eigenvalues and such that all its eigenstates are orthonormal and, moreover, form a complete basis, then this operator is Hermitian. This is because you can pick its eigenbasis as an orthonormal basis and in that basis it will be represented by a manifestly ...


2

The spectral theorem only holds for normal operators. Self-adjoint operators are normal, symmetric ones not necessarily so. In physicist-speak, we want the generalized eigenvectors to from a 'complete basis' of the Hilbert space. For example, the generalized eigenvectors of the momentum operator in position representation are plane waves, and even though ...


2

I'm pretty sure there exists an answer for this here already, but I can't find it (it's always about unboundedness). For the Hamiltonian, the answer is basically given by Stone's theorem on one-parameter unitary groups. There is a one-to-one correspondence between self-adjoint operators and strongly continuous one-parameter families of unitaries. Why is ...


3

Warning: This post contains a really stupid but important error. I will fix it in a couple of hours, but for now please don't read it. There's a bit more to this than the other answers are covering. As you already noted, in expressions like this $$(c_1^\dagger c_2^\dagger \ldots) |0\rangle,$$ it is important to keep a consistent ordering. The reason is ...


2

I would interpret $1$ and $2$ in the function arguments as shorthands for the coordinates belonging to particles 1 and 2; but the subscripts $1, 2, i, k$ of the single-particle wave functions $\phi$ / $\psi$ and corresponding creation/annihilation operators $a^+$ / $a$ as specifying a single-particle state. The first part is quite a common convention: ...


2

As OP pointed out, one should stick to a single convention. Hence the order of the creation operators should be reversed in one of the two cited equations. Reversing the order of operators in which of the equations is purely a matter of choice. However, to obtain the simplest relation between the Slater determinant and the occupation number representations, ...


3

Since the states $|\psi\rangle$ and $-|\psi\rangle$ differ only by a phase factor, they really describe the same electron configuration. So, while it is a bit sloppy to have inconsistent sign conventions in different places in a text, it's not surprising to see it, and it's not really a big deal. (Unless you have inconsistent sign conventions within a single ...


1

Consider an operator $A$ on a Hilbert space $\cal H$, say $L^2(\mathbb R)$ in order to deal with QM of a particle on a real axis without spin. Let $D(A) \subset \cal H$ be the domain of $A$. The spectrum $\sigma(A)\subset \mathbb C$ of $A$ is defined as the union of the following three pairwise disjoint subsets $\sigma_p(A)$, $\sigma_c(A)$, $\sigma_c(A)$. ...


0

What you did seems fine to me, and indeed you can show that $\frac{\partial}{\partial{p}^k}(a_\vec{p}^\dagger{a}_\vec{p})=0$, just take the definitions $$a_\vec{p}=\int{d^3x}\,e^{-i\vec{p}\cdot\vec{x}}\left[\frac{E_\vec{p}}{2}\phi(\vec{x})+\frac{i}{E_\vec{p}}\dot\phi(\vec{x})\right]\\ ...


3

Any Hilbert space $\mathcal{H}$ with the notion of unitary time evolution also possesses the notion of Hamiltonian. If $\mathcal{U}(t) : \mathcal{H} \to \mathcal{H}$ is the time evolution operator for every $t \in \mathbb{R}$, then it forms a one-parameter Lie subgroup of the Lie group of unitary operators, which is generated by some distinct element $H$ ...


4

No, I don't see why that should be the case. The notion of a Hilbertspace underlying a quantum-mechanical system is quite independant of the postulate that time evolution is generated by a Hamiltonian. The notion of a vectorspace enters QM, because fundamentaly QM should be a linear theory and thus allow for arbitrary superpositions. The more wonderous ...


3

It seems, the state $|\psi\rangle$ is a superposition of $|\phi\rangle$ and several eigenstates of the Hamiltonian: $$ \hat{H}|\chi_n\rangle = E_n|\chi_n\rangle $$ The sigma then just denotes the sum of the eigenstates. And since the Hamilton operator is linear, you can easily apply it to each element of the sum independently. $$ \hat{H}|\psi_n\rangle = ...


2

Let $\mathcal{H}$ be the Hilbert space for one particle. Then, $S_{x}\in\mathcal{B}(\mathcal{H})$ is a bounded, self-adjoint operator. Now, if you want to have the Hilbert space for two particles, remember that this is the tensor product, i.e. $\mathcal{H}=\mathcal{H}_1\otimes \mathcal{H}_2$ (where $\mathcal{H}_1$ is the Hilbert space of the first and ...


1

Hermitian observables yield real eigenvalues. Also, for an observable to be a good observable, it must be diagonal in some basis, in which case it has real diagonal elements. So your observable is related to a real, diagonal matrix by a unitary transformation. That pretty much restricts what you can have. This isn't axiomatic, of course. It's more from a ...


0

Ok, there are a lot of points here. 1) First of all, an operator in Hilbert spaces is not defined only by its action (e.g. the operation of derivation for the momentum), but also by the so-called domain of definition, i.e. the subspace of vectors of the Hilbert space where it can act. Unbounded operators are not defined for every vector of the Hilbert ...


2

If you have different Hilbert spaces, you cannot say it is the same operator on them, since operators are defined on the Hilbert space. The momentum operator is a tricky one for many systems, and rigor requires the discussion of concepts like rigged Hilbert spaces. A nice introductory discussion of this is "Mathematical surprises and Dirac's formalism in ...


0

Time reversal is not only complex conjugate, what it does is also to transpose the items on which it acts (vectors, matrices). T<ϕ|Ô|ψ> = <ψT|Ô|Tϕ>. Notice the change of places of the functions in the right wing with respect to the left wing. Also, I used the fact that Ô is unchanged at time-reversal. Now we do the following change which is allowed ...


2

This is quite simple. Consider the operator $H$ on the Hilbert space $\mathscr{H}$, in your simple example it has a spectral resolution: $$H=\sum_{n}E_n \lvert n\rangle\langle n \rvert\; .$$ Each eigenvalue has multiplicity 1. Now the operators $H_1$ and $H_2$ on $\mathscr{H}\otimes\mathscr{H}$ have the same spectrum of $H$, but each eigenvalue has ...


2

Quantum dynamics is commonly known to be generated by self-adjoint operators. Therefore in order to properly define the dynamics of a system it is necessary to introduce a suitable self-adjoint Hamiltonian operator. In quantum field theories, this task is extremely difficult, because the formal operators that emerge quantizing a classical field theory ...


2

The first equation is not normalized, which is the result of the lowering operator. While the second equation is the normalized state, you can check it easily by using $\langle 10|10 \rangle$. Or you can use the normalization condition $\langle \psi|\psi \rangle = 1$ where $|\psi \rangle = Z \hbar | \uparrow \downarrow + \downarrow \uparrow \rangle$.


2

If one calculates expectation value $\langle\Psi| M^2|\Psi\rangle$ of the mass-square operator $M^2$ of the state $|\Psi\rangle~=~ \alpha^j_{-1}| 0; p\rangle $, then the level-matching condition (2.25) would be violated.


3

Define $$ N \equiv \sum_{i=0}^{D-2} \alpha_{-n}^i \alpha_n^i,~~~~{\tilde N} \equiv \sum_{i=0}^{D-2} {\tilde \alpha}_{-n}^i {\tilde \alpha}_n^i $$ The formula you have written tells us un particular that for any state, we must have $N = {\tilde N}$. This condition is known as the level-matching condition. The state $\alpha^i_{-1} |0;k\rangle$ has $N = 1$ but ...


2

Let's take the spin, it's the simplest case, $Q = \mathbf{s}$. The operator $\mathbf{s}$ is a vector, \begin{equation}s = \mathbf{i}s_x + \mathbf{j}s_y + \mathbf{k}s_z\end{equation} while the operator s^2 is a scalar \begin{equation}s^2 = s_x^2 + s_y^2 + s_z^2\end{equation}. The operators $s_x$, $s_y$, and $s_z$ don't commute two by two, but $s_x^2$, ...


1

The Hilbert space of the states in this case is the Fock space. It is a linear space "constructed" by acting by the creation operators $a^\dagger_k$ on the vacuum state $|0\rangle$, which has the property $a_k|0\rangle=0$. All other states are related so that the commutation relations between $a,a^\dagger$ are satisfied. The individual states like ...


2

Usually in many body theory these operators create and annihilate particles. There are different annihilation and creation operators for fermions and bosons (they obey different commutation relations). The states they act upon and the states "created" by them respect the required symmetries (antisymmetric for fermions, symmetric for bosons). The operators ...



Top 50 recent answers are included