quantum mechanics - Energy operator

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.

Answer

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.

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  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.

  
  


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  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.