# Proper notation and definition of the Hamilton operator

The Hamilton operator is often defined as $$\hat H = \frac{-\hbar^2 }{2m}\frac{d^2}{dx^2} + V(x)$$ but shouldn't it rather be \begin{aligned} \hat H &= \int\int dxdx' |x\rangle \langle x|\hat H|x'\rangle \langle x'|\\ &= \int \int dx dx' |x\rangle \left[ -\frac{\hbar^2}{2m}\left(\frac{\partial}{\partial x'}\frac{\partial}{\partial x}\delta(x-x') \right) + V(x)\delta(x-x') \right ]\langle x'| \\ \hat H &= \int dx |x\rangle \left[ -\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2} + V(x) \right ]\langle x| \\ \hat H &= \int dx |x\rangle \langle x| \left[ -\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2} + V(x) \right ]\\ \hat H&=\hat 1\left[ -\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2} + V(x) \right ] \end{aligned}

Because it seems to me that only in this way we have

$$\hat H |\psi\rangle = \int dx |x\rangle \left[ -\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2} + V(x) \right ]\psi(x)$$ and $$\langle x|\hat H | \psi\rangle =\left[ -\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2} + V(x) \right ]\psi(x)$$

Am I overly pedantic or even wrong? Is the distinction between $$\hat 1 \left[ -\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2} + V(x) \right ] = \hat H$$ and $$\left[ -\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2} + V(x) \right ]$$ unnecessary/unreasonable or should it be made?

• Consider to number your equation as it is hard to refer to a particular one. The step to your equation before 'Because it seems...' is not correct. Jul 22, 2022 at 14:01
• You may be interested in my answer to a related question here. Jul 22, 2022 at 14:58
• Your sign is wrong, bro. $p^2 \to -\hbar^2\nabla^2$. It is $-\nabla^2$ that is actually the non-negative definite operator. Besides that, yes, you are being a little bit pedantic, IMHO. But don't worry about it too much because you will find a lot of even bigger pedants in the field of physics.
– hft
Jul 22, 2022 at 16:46
• @hft Thank you for pointing the sign error out. It went over my head while typing everything else. Jul 23, 2022 at 8:26
• @J.Murray Your link answers my question. I'd wish the distinction was made as clearly as in your post, then my question would be superfluous and it would have spared me a lot of confusion. Jul 23, 2022 at 8:34

You are not pedantic, you are just wrong! $$\hat H= \hat p^2 /2m+ V(\hat x).$$ Recall $$\hat p= -i\hbar \int\! dx ~|x\rangle \partial_x \langle x| ,\\ \hat x= \int\! dx ~|x\rangle x \langle x|~, \qquad \langle x| x'\rangle = \delta (x-x'),\\ \leadsto \qquad \hat p^2= -\hbar^2 \int\! dx ~|x\rangle \partial_x^2 \langle x| ,$$ so that \begin{aligned} \hat H &= \int\int\! dxdx' ~|x\rangle \langle x|\hat H|x'\rangle \langle x'|\\ &= \int \int\! dx dx' ~|x\rangle \delta(x-x') \left( - \frac{\hbar^2}{2m} \partial_{ x'}^2 + V(x') \right )\langle x'| \\ &= \int\! dx ~~|x\rangle \left ( -\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2} + V(x) \right )\langle x| ~~. \end{aligned}

Your teacher surely has taught you that the operator in the parenthesis is the hamiltonian in the x-representation! Thus, acting on a ket, $$\hat H |\psi\rangle = \int\! dx ~~|x\rangle \left ( -\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2} + V(x) \right ) \psi(x) ~~.$$

• I assure you that my teachers never taught me this. Rather the opposite, they caused my confusion. I am trying to relearn things the proper way, which is often different to the way it was presented to me. Did you take a look at the wikipedia site that I linked? If I understand you correctly, the article is wrong. Just take a look at the One particle section. Jul 22, 2022 at 17:26
• @HansWurst Wikipedia is not wrong per se. As I've tried to state in my answer, the Hamiltonian as given there (or in my answer the first equation) is fine. It is just an operator defined for a particular Hilbert space, namely $L^2(\mathbb R)$, where the position operator is multiplicative and the momentum operator acts as a derivative operator (a.k.a. position representation). Jul 22, 2022 at 17:29
• @HansWurst WP is sloppy on this, in a time-honored physics way, confident the reader has been introduced to QM. Instead of an equal sign, I'd use and arrow, or a qualifier "and, in the coordinate representation = " or something of the sort. It is not a big deal. The expression you must know and use is $\hat p$ in the coordinate representation, and the appreciation of how that presents in the bra-ket notation; Dirac, with a mathematical and engineering background, invented it expressly to prevent such confusions... Jul 22, 2022 at 17:37

The operator (as you've used it in your first equation)

$$H=- \frac{\mathrm d^2}{\mathrm dx^2} + V(x)\tag{1}$$

is defined on (a subspace of) $$L^2(\mathbb R)$$, whereas the Dirac notation does not make an explicit reference to a particular Hilbert space, i.e. it is some complex separable Hilbert space of infinite dimension and thus isomorphic to $$L^2(\mathbb R)$$. For example, when we write

$$H=P^2+ V(X) \tag{2} \quad ,$$ we a priori do not specify on which Hilbert space these operators ($$H$$, $$P$$ and $$X$$) act on or how they act on the respective vectors (think about position versus momentum representation). To quote B. Hall. Quantum Theory for Mathematicians. Springer. Section 3.12.:

One peculiarity of the physics literature on quantum mechanics is a conspicuous failure of most articles to state what the Hilbert space is. Rather than starting by defining the Hilbert space in which they are working, physicists generally start by writing down the commutation relations that hold among various operators on the space. Thus, for example, a physicist might begin with position and momentum operators $$X$$ and $$P$$, satisfying $$[X,P ] = i\hbar I$$, without ever specifying what space these operators are operating on. [...] It is, nevertheless, disconcerting for a mathematician to encounter an entire paper full of computations involving certain operators, without any specification of what space these operators are operating on, let alone how the operators act on the space.

• yeah but most of the time the Hilbert space is clear from context (explicit or historical), and likewise for the action of the operators. Most of the time, math people have their own idiosyncrasies, as anyone who has ever read a grant proposal in math can attest. Jul 22, 2022 at 14:57
• Indeed, physicists assume a common culture and good faith; mathematicians are trained to look for cracks and nicks, while physicists assume good faith and rush to the point. As Gell-Mann is known to have told his collaborator, Bruno Renner, "Brunno, if we put that stuff in we'll confuse somebody; as things stand, any intelligent, well-meaning reader will know exactly what we mean!". He was, of course, epically right. Jul 22, 2022 at 17:05