I am trying to learn Quantum mechanics from MIT OCW Videos about quantum mechanics. I have reached the 5th lecture. Please help me in understanding this:

In the middle (At 32:08), the professor wrote that the $$\displaystyle\text{Energy Operator}={\hat p^2\over2m}+V(\hat x).$$


  1. From where do we get this equation?
  2. What is $V(\hat x)$ here?
  3. Is $V(\hat x)\overset?=V(x)$?
  4. Afterwards (At 1:15:46), the professor wrote $$\hat E=i\hbar{\partial\over \partial t}.$$ So are there two energy operators?
  • $\begingroup$ How familiar are you with Classical Mechanics, specifically the Hamiltonian formulation? $\endgroup$
    – Kyle Kanos
    Jun 28, 2014 at 15:00
  • $\begingroup$ @KyleKanos I have never heard of it. I am not very familiar with QM. I have just started learning it. $\endgroup$
    – Kartik
    Jun 28, 2014 at 15:03
  • 2
    $\begingroup$ I would recommend pausing your learning QM for a little bit and focus on reading up on the Hamiltonian formalism of classical mechanics. $\endgroup$
    – Kyle Kanos
    Jun 28, 2014 at 15:04
  • $\begingroup$ @KyleKanos OK. Can you tell me that how can I learn this? $\endgroup$
    – Kartik
    Jun 28, 2014 at 15:05
  • $\begingroup$ I believe MIT OCW has a Classical Mechanics. I know for sure that Stanford's Online coursework has a Classical Mechanics series that you can use. $\endgroup$
    – Kyle Kanos
    Jun 28, 2014 at 15:09

2 Answers 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 writing $\frac{1}{2}mv^2$) and $V(x)$ is the potential energy.

$V$ is first of all just a function of its argument. If you write $V(x)$, you evaluate this function at (the number) $x$. If you write $V(\hat x)$, you replace the position (number) with the position operator, so the whole thing, $V(\hat x)$ is also an operator, specifically the potential energy operator.

This is how you obtain the energy operator $\hat E$ (also called Hamiltonian and thus conventionally written as $\hat H$) from the correspondence principle. This is more of an axiom of quantum mechanics, there is no inherent motivation. The idea is that in the classical limit, the results of classical physics, specifically the classical expression for the energy, should be retrievable.

Now, when you want a specific realization of the position and momentum operators on the Hilbert space the wave functions are going to live in, you replace $\hat x\psi(x)$ by $x\psi(x)$, i.e. the position operator acts on a wave function by multiplying it with $x$, and $\hat p\psi(x)$ with $-i\hbar\partial_x\psi(x)$. By the same token, the energy operator is written as $i\hbar\partial_t$. Note, that the form of the momentum operator and the energy operator are somewhat similar they differ only by $\partial_x$ being replaced with $\partial_t$. If you know Noether's theorem, you will be able to appreciate this fact.

Equating both forms of the energy operator gives the time-dependent Schrödinger equation: $$i\hbar\partial_t\psi(x)=\hat H\psi(x)$$ where $$\hat H = \frac{\hat{p}^2}{2m}+V(\hat x)$$


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 coordinates $x$ and $p$ to operators fulfilling the canonical commutation relation $[\hat x,\hat p] = \mathrm{i}\hbar$, we can obtain the Hamiltonian operator $\hat H$ from the classical $H$ by replacing the phase space coordinates by their operator versions, this gives the $\hat H$ you give first.

The relation to time evolution comes about since, classically, the time evolution of any observable $f$ is given by $\frac{\mathrm{d}f}{\mathrm{d}t} = \{f,H\}$, so the Hamiltonian is the generator of the time translation. Transferring this to the quantum theory means that the Hamiltonian of any system will also generate the time translations in the quantum theory, this is encapsulated in the Schrödinger equation

$$ \mathrm{i}\hbar\partial_t|\psi(t)\rangle = \hat H|\psi(t)\rangle $$

which arises as the Hamilton-Jacobi equation of the quantum theory. It is important to note that $\hat H = \mathrm{i}\hbar\partial_t$ is abuse of notation (at least as I see it), since $\partial_t$ is not a proper operator on the space of states.

(As an aside, I would advise anyone learning quantum mechanics by the canonical approach to first get a firm grasp on the concepts of classical Hamiltonian mechanics, since it heavily draws from analogies to that.)


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