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Does the Hamiltonian always translate to the energy of a system? What about in QM? So by the Schrodinger equation, is it true then that $i\hbar{\partial\over\partial t}|\psi\rangle=H|\psi\rangle$ means that $i\hbar{\partial\over\partial t}$ is also an energy operator? How can we interpret this? Thanks.

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For the first part of the question (v1), see also physics.stackexchange.com/q/11905/2451 For the second part of the question (v1), see also physics.stackexchange.com/q/17477/2451 –  Qmechanic Oct 13 '11 at 18:33

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up vote 4 down vote accepted

I will formulate the following in such a way, that the language doesn't change too much within the answer. This also emphasizes the analogies of related concepts.

  • Classically, you have a configuration/state $\Psi$, which is characterised by coordinates $x^i,v^i$ or $q^i,p_i$ and/or any other relevant parameters. Then an energy is a function or functional of this configuration

    $$H:\Psi\mapsto E_\Psi,\ \ \mbox{where}\ \ E_\Psi:=H[\Psi].$$

    Here $E_\Psi$ is some real (energy-)value associated with the configuration $\Psi$.

    To name an example: Let $q$ and $p$ be the coordinates of your two-dimensional phase space, then every point $\Psi=(q,p)$ characterises a possible configuration. The configuration/state $\Psi$ here is really just the pair of coordinates. The scalar function $H(p,q)=\frac{1}{2m}p^2+\frac{\omega}{2}x^2$ clearly is a map which assigns a scalar energy value $E_\Psi$ to every possible configuration $\Psi$.

    The evolution of $\Psi$ in time is determined by $H$, see Hamilton's Equations. This might be viewed as the point of coming up with the Hamiltonian in the first place and it is typically done in such a way, that the energy value $E_\Psi$ will not change with time. See also this thread for a related question. What you call "energy" is pretty much determined by this criterium. In the case of a time independent Hamiltonian (as in the example) and if the time developement of observables $f$ is governed by $\frac{\mathrm{d}f}{\mathrm{d}t} = \{f, H\} + \frac{\partial f}{\partial t}$, then you have $\frac{\mathrm{d}H}{\mathrm{d}t} = \{H, H\} = 0$ and the conservation of the quantiy $E_\Psi:=H[\Psi]$ is evident. Of course, you might want to model friction processes and whatnot and it then might be difficult to define all the relevant quantities.

  • In quantum mechanics, your configuration $\Psi$ is given by a state vector $|\Psi\rangle$ (or an equivalence class of such vectors) in some Hilbert space. There are many vectors in this Hilbert space, but there are some vectors $|\Psi_n\rangle$, which also span the whole vector space and which are also special in the following sense: They are eigenvectors of the Hamiltonian operator: $H|\Psi_n\rangle = E_n|\Psi_n\rangle$. Here $E_n$ is just the real eigenvarlue and I assume that I can enumerate the eigenstates by an descrete index $n$. Now for every point in time, your state vector $\Psi$ is just a linear combination of the special states $\{\Psi_n\}$. (As a remark, notice that all the time dependencies of states are left implicit in this post.) Therefore, if you know how $H$ acts on all the $\Psi_n$'s, you know how $H$ acts on any $\Psi$. Since a Hilbert space naturally comes with an inner product, i.e. a map

    $$\omega:|\Psi\rangle\times|\Phi\rangle\mapsto\omega(|\Psi\rangle,|\Phi\rangle)\equiv\langle\Psi|\Phi\rangle\in\mathbb{C},\ \ \mbox{satisfying}\ \ \langle\Psi|\Psi\rangle>0\ \ \forall\ \ |\Psi\rangle\ne 0,$$

    you can define a new map

    $$\omega_H:\Psi\mapsto E_\Psi,\ \ \mbox{where}\ \ E_\Psi:=\omega_H[\Psi],$$

    with

    $$\omega_H[\Psi]:=\omega(|\Psi\rangle,H|\Psi\rangle)\equiv\langle\Psi| H|\Psi\rangle.$$

    Compare the lines above with the classical case. Here $E_\Psi=\ ...=\langle\Psi| H|\Psi\rangle$ is then called the expectation value of the Hamiltonian in the phyical state. It is the energy value associated with $\Psi$, which is real due to hermiticity of the Hamiltonian. Also, like in the classical case, the time evolution of any state $\Psi$ (resp. state vector $|\Psi\rangle$) is determined by the observable $H$, an operator in the QM-case. And as stated above, exactly this $H$, together with the state/configuration $\Psi$, gives you the energy values $E_\Psi$ associated with $\Psi$. This relation of time and energy is by construction: The Schrödinger equation is an axiom (but a natural one, see conservation of probability), which relates time evolution and Hamiltonian. Now, if the time dependency of the state is governed by the Hamiltonian (whatever it might look like in your scenario), then so is the time dependency of $\langle\Psi| H|\Psi\rangle$.

    And if $\ i\hbar\frac{\partial}{\partial t}|\Psi\rangle=H|\Psi\rangle\ $ is true for all vectors in your Hilbert space, i.e. if $i\hbar\frac{\partial}{\partial t}=H$ holds as an operator equation, then these two really are just the same operator. If you ask for an interpretation for this, then I'd suggest you hold on to the quantum mechanical relation between frequency and energy. Regarding the equation which determines time evolution, quantum mechanics is much easier than classical mechanics in a sense, especially if you come with some Lie group theory intuition in your backpack.

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The classical example for something where the Hamiltonian is different from the total energy is a particle in an accelerating constraint, like a particle bead sliding on a rotating wire. I will use a different system, a particle of mass m in a long uniformly accelerating box.

If the box is accelerating with acceleration a, in the comoving system, there is a fictitious force on the particle which is derived from a fictitious potential. The comoving Hamiltonian description is the same as for a particle in gravity, so that

$$ H = {p^2\over 2m} + mg x$$

Which is valid for positive x, and the potential is infinite for negative x. Viewing the same particle in the non-accelerated frame, the total energy is just the kinetic energy, and the potential energy restricts the particle from entering the region $x<{at^2\over 2}$. The comoving Hamiltonian is not the energy of the particle, which increases without bound with time, but it gives the dynamical law for the comoving frame wavefunction.

The wavefunction of the particle will (if it can radiate) settle down to the ground state of the moving Hamiltonian. The particle will be in a bound profile against the wall, where the binding is by a linear potential. For the inertial frame, this profile will be accelerating steadily, and its energy does not settle down. The relation between the two is given by boosting the wavefunction by an amount which depends on time.

For systems which are not constrained, the Hamiltonian is always the total energy. This is also true for systems where the constraints do not add energy to the system. The Hamiltonian for systems which add energy is usually explicitly time dependent, but not so in the case where the dynamics is time independent from the point of view of the particle. Mathematically, in such a system you have a nontrivial time translation invariance which is a symmetry, and in the accelerated particle case, this time translation symmetry mixes up inertial frame time translation and boosts.

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But time is not regarded as an operator but as a parameter in quantum mechanics. Right? Then, is the replacement $E\rightarrow i\hbar\frac{\partial}{\partial t}$ valid? If yes, I would like to know whether it is hermitian. –  Roopam Feb 21 at 17:44

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