64 votes
Accepted

Why do excited states decay if they are eigenstates of Hamiltonian and should not change in time?

The atomic orbitals are eigenstates of the Hamiltonian $$ H_0(\boldsymbol P,\boldsymbol R)=\frac{\boldsymbol P^2}{2m}+\frac{e}{R} $$ On the other hand, the Hamiltonian of Nature is not $H_0$: there ...
AccidentalFourierTransform's user avatar
28 votes
Accepted

On Groenewold's Theorem and Classical and Quantum Hamiltonians

You probably need to internalize Ivan Todorov's accessible Quantization is a mystery. Your best bet for addressing your questions is Geometric quantization, not phase space quantization that you ...
Cosmas Zachos's user avatar
27 votes
Accepted

Is it possible to reconstruct the Hamiltonian from knowledge of its ground state wave function?

IF you know that your Hamiltonian is of the form $$ \hat H=\frac{-\hbar^2}{2m}\nabla^2+V(\mathbf r) \tag 1 $$ for a single massive, spinless particle, then yes, you can reconstruct the potential and ...
Emilio Pisanty's user avatar
27 votes

Why do excited states decay if they are eigenstates of Hamiltonian and should not change in time?

The hydrogen atom in an excited state is not really in an energy eigenstate. There are two ways of looking at it. One way is to recognize that the atom is not isolated. It is always coupled to ...
garyp's user avatar
  • 22.2k
27 votes

The position-representation matrix elements of the propagator for a particle in a ring

You wouldn't think it, from how easy it is to pose this question, but it is ridiculously nontrivial. As it happens, it is entirely impossible to find the position-basis matrix elements of this ...
Emilio Pisanty's user avatar
27 votes
Accepted

Why is the ground state important in condensed matter physics?

To add to Vadim's answer, the ground state is interesting because it tells us what the system will do at low temperature, where the quantum effects are usually strongest (which is why you're bothering ...
taciteloquence's user avatar
25 votes

Why symmetry transformations have to commute with Hamiltonian?

Sometimes this is claimed without much explanation. The time evolution operator is given by exponentiating the Hamiltonian: $$ U(t) = \exp(-i t\hat H / \hbar ). $$ For concreteness, when we think ...
user1379857's user avatar
  • 11.5k
24 votes
Accepted

How do I simulate an atom?

The Hamiltonian for the He atom is: $$H = -\frac{\hbar^2}{2m_e}(\nabla_1^2 + \nabla_2^2) - \frac{2e^2}{4\pi\epsilon_0 r_1} - \frac{2e^2}{4\pi\epsilon_0 r_2} + \frac{e^2}{4\pi\epsilon_0 r_{12}}$$ ...
Anders Sandberg's user avatar
23 votes
Accepted

Where does the $i$ come from in the Schrödinger equation?

Let $U$ be an unitary operator. Write $$ U=\mathbb I+\epsilon A $$ for some $\epsilon\in\mathbb C$, and some operator $A$. Unitarity means $U^\dagger U=\mathbb I$, i.e., $$ U^\dagger U=(\mathbb I+\...
AccidentalFourierTransform's user avatar
21 votes

Detailed derivation and explanation of the AKLT Hamiltonian

Let me try to answer my own questions to thank those who gave voice and support for undeleting this question. My main reference is the chapter 3 Basic quantum statistical mechanics of spin systems of ...
Cheng Guo's user avatar
  • 609
20 votes
Accepted

In the quantum hamiltonian, why does kinetic energy turn into an operator while potential doesn't?

Actually the potential is also an operator. It just so happens that, in the position representation, $\hat x\psi(x)=x\psi(x)$, so that the potential energy operator $V(\hat x)$ acts by multiplication:...
ZeroTheHero's user avatar
  • 45.5k
19 votes
Accepted

What is the Hamiltonian of General Relativity?

One should be careful to distinguish the Hamiltonian associated with particle motion in a gravitational field and the Hamiltonian associated with the gravitational field itself. The theorem on page ...
gj255's user avatar
  • 6,375
19 votes

Momentum operator acting on a bound state doesn't return an eigenvalue although kinetic energy operator does. Why?

For the specific case of the particle-in-a-box problem, the momentum operator is much trickier to handle than what you're allowing for, but that's not the issue at the heart of your current confusion. ...
Emilio Pisanty's user avatar
19 votes
Accepted

Why particle hole symmetry and chiral symmetry are called symmetries?

They are called symmetries because (when the symmetry exists) they commute with the second quantized Hamiltonian: $$\hat{H} = \sum_{AB}\hat{\psi}^{\dagger}_A H_{AB} \hat{\psi}_B,$$ where $H_{AB}$ are ...
David Bar Moshe's user avatar
19 votes
Accepted

Symmetry in quantum mechanics

In a context like this, a symmetry is a transformation that converts solutions of the equation(s) of motion to other solutions of the equation(s) of motion. In this case, the equation of motion is ...
Chiral Anomaly's user avatar
17 votes
Accepted

What is the Hamiltonian of a Measurement?

The following argument is very vague and hand-wavey - as is inevitable when discussing measurement apparatus with on the order of $10^{23}$ particles. It is intended only to illustrate the idea. We ...
Nullius in Verba's user avatar
17 votes
Accepted

If the Lagrangian depends explicitly on time then the Hamiltonian is not conserved?

"how can you be sure there does not exist a Lagrangian depending explicitly on time such that the partial derivative with respect to time vanishes for all 𝑡 " If the question is that, it ...
Valter Moretti's user avatar
16 votes

Dealing with tensor products in an exponent

I just found this post because I was confused by the same step. But I think I got it now with the help of @lionelbrits post and @Chris2807's comment. Just adding this for completeness and maybe ...
Tera's user avatar
  • 502
16 votes

What is meant by unitary time evolution?

Yes, there is a difference. Unitary time evolution is the specific type of time evolution where probability is conserved. In quantum mechanics, one typically deals with unitary time evolution. ...
Avantgarde's user avatar
  • 3,832
16 votes

Do non-commuting Hamiltonians have non-commuting time evolution operator?

Here is a counterexample to the first part of OP's question: Imagine $K$ is diagonalizable with eigenvalue spectrum $\subseteq \mathbb{Z}$ within the integers. Then $e^{i2\pi K}={\bf 1}$ is the ...
Qmechanic's user avatar
  • 202k
15 votes
Accepted

How can a Hamiltonian determine the Hilbert space?

The point is that sometimes one starts from a more or less explicit algebraic formalism where only algebraic manipulations of algebra elements are initially used. Here operators are not operators in a ...
Valter Moretti's user avatar
15 votes

In the quantum hamiltonian, why does kinetic energy turn into an operator while potential doesn't?

OP's hunch is exactly right: In QM they are all operators. However each operator may become a multiplication operator in certain representations. E.g. position operators become multiplication ...
Qmechanic's user avatar
  • 202k
15 votes
Accepted

Why I cannot write the time evolution operator $e^{-i(T+V)t}$ as the product of operators $e^{-iTt}e^{-iVt}$

This is because of the BCH formula $$\begin{align}e^Z~=~&e^Xe^Y \cr\Downarrow~&\cr Z~=~&X+Y+\frac{1}{2}[X,Y]+{\cal O}(X^2Y,XY^2),\end{align}$$ or equivalently, the Zassenhaus formula. But ...
Qmechanic's user avatar
  • 202k
15 votes
Accepted

Property of the Hamiltonian's discrete spectrum

This is the so-called Weinstein criterion.$^1$ To prove it, we proceed as OP and expand the (normalized) vector $\psi$ into the (assumed to be) complete orthonormal eigenbasis of $H$ and denote by $...
Tobias Fünke's user avatar
14 votes

Why is the ground state important in condensed matter physics?

Ground state contains information about most thermodynamic properties of the system at zero temperature. In fact, it can be thought of as a limiting case of the partition function at zero temperature. ...
Roger V.'s user avatar
  • 58.7k
13 votes

The formal solution of the time-dependent Schrödinger equation

The existing answer by Qmechanic is entirely correct and extremely thorough. But it is very long and technical, and there's a danger that the core of the answer can get buried under all of that. The ...
Emilio Pisanty's user avatar
13 votes
Accepted

Can we make sense of a Hamiltonian $a^\dagger a^\dagger + a a$?

No, the ground state is not well-defined because the energy is unbounded below. To see this, switch back to the variables $x$ and $p$ using $a \sim x + ip$ to find $$H \sim p^2 - x^2.$$ This is the ...
knzhou's user avatar
  • 102k
13 votes
Accepted

Why are Hamiltonian Mechanics well-defined?

This is a good but quite broad question. Let us suppress position dependence $q^i$, $i\in\{1, \ldots, n\}$, and explicit time dependence $t$ in the following to keep the notation simple. Given a ...
Qmechanic's user avatar
  • 202k
13 votes
Accepted

A confusion about the spin of a particle described by Dirac equation

It is true that the Dirac Hamiltonian does not commute with the spin vector $\vec{S}$. While we have $\left[\vec{J},H_{D}\right]=0$, the related commutator with $\vec{S}$ is nonzero. So the vector ...
Buzz's user avatar
  • 16.1k
12 votes
Accepted

Bogoliubov-de-Gennes (BdG) formalism of Hamiltonians

The Bogoliubov-de-Gennes Hamiltonian is a mean-field Hamiltonian, that is, a one-body (quadratic) Hamiltonian: it is by no means equivalent to the original many-body Hamiltonian. The two-body forces ...
C. Berthod's user avatar

Only top scored, non community-wiki answers of a minimum length are eligible