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For which systems we represent the Hamiltonian as a differential operator and for which system we represent it by a matrix? Can the momentum be represented by a matrix operator?

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  • $\begingroup$ You might be also interested in physics.stackexchange.com/a/90078/26076 where I look at the QHO and show that we can have both kinds of operators "matrix" or "differential", depending on the co-ordinates we choose. $\endgroup$ – WetSavannaAnimal Feb 5 '14 at 0:16
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If the Hilbert space of the system in question is finite-dimensional, then in a given basis for the Hilbert space, the Hamiltonian (and every other observable for that matter), will be represented by a matrix.

If the Hilbert space is infinite-dimensional, the situation is a bit different. In Quantum Mechanics, we typically assume that the Hilbert spaces we deal with are separable which means that they admit a countable, orthonormal basis. The representation of the Hamiltonian in any such basis will be a "matrix" that is "infinite-dimensional."

Consider, for example, the quantum harmonic oscillator. Let $B = \{|n\rangle\}$ denote the energy eigenbasis where $|n\rangle$ is the energy eigenvector with eigenvalue $E_n = (n+\frac{1}{2})\hbar\omega$, then the Hamiltonian in this basis looks as follows: \begin{align} [\hat H]_B = \begin{pmatrix} E_0 & & & \\ & E_1 & & \\ & & E_2 & \\ & & & \ddots\\ \end{pmatrix} \end{align} Moreover, the momentum operator is also represented as an "infinite-dimensional" matrix in this basis. Recall that the momentum operator can be written in terms of creation and annihilation operators as follows: \begin{align} \hat p = i\sqrt{\frac{m\omega\hbar}{2}}(\hat a^\dagger - \hat a) \end{align} which gives \begin{align} \langle m|\hat p|n\rangle &= i\sqrt{\frac{m\omega\hbar}{2}}(\sqrt{n+1}\langle m|n+1\rangle - \sqrt{n}\langle m|n-1\rangle) \\ &= i\sqrt{\frac{m\omega\hbar}{2}}(\sqrt{n+1}\delta_{m,n+1} - \sqrt{n}\delta_{m,n-1}) \end{align} and we can therefore easily write out the first few entries of the momentum operator written in the energy eigenbasis: \begin{align} [\hat p]_B = i\sqrt{\frac{m\omega\hbar}{2}}\left( \begin{array}{ccccc} 0 & -1 & & & \\ 1 & 0 & -\sqrt{2} & & \\ & \sqrt{2} & 0 & -\sqrt{3} & \\ & & \sqrt{3} & 0 & \ddots \\ & & & \ddots & \ddots \\ \end{array} \right) \end{align} On the other hand, every state in the Hilbert space can be written in the so-called position $\{|x\rangle\}$ and momentum $\{|p\rangle\}$ "bases." These aren't strictly speaking bases of the Hilbert space, but they pretty much work the same way in the sense that a given state can be written as a integral over these "continuous" basis elements as opposed to sums over "discrete" bases; \begin{align} |\psi\rangle &= \int_{\mathbb R}dx\psi(x)|x\rangle, \\ |\psi\rangle &= \int_{\mathbb R}dp\tilde\psi(p)|p\rangle \end{align} In this notation, every physical state corresponds to a square integrable function $\psi$ in the position basis and $\tilde\psi$ in the momentum basis. It is in these "bases" that the various observables are represented by multiplication/differential operators on a function space. In particular, for example the momentum operator is represented as follows: \begin{align} (\hat p\psi)(x) &= \langle x|\hat p|\psi\rangle = \frac{\hbar}{i}\frac{d\psi}{dx}(x) \\ (\hat p\tilde\psi)(p) &= \langle p|\hat p|\psi\rangle = p\tilde \psi(p). \end{align} In short, the representation of observables, whether they are matrix or differential operator representations, depends on the basis you decide to represent them in.

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    $\begingroup$ @joshphysics- A spin half system, for example, has a finite dimensional hilbert space. Right? We can choose, $S_z$ basis or energy eigenbasis to explicitly write down the $2\times 2$ matrix representation of the operators in this case. Right? For this particular system, is there a concept of position or momentum representation? If yes, how do these operators look? Also I am having the feeling that, momentum operator written in momentum basis or position operator written in position basis can't have matrix representation. Am I right? $\endgroup$ – SRS Feb 4 '14 at 3:35
  • $\begingroup$ @RoopamSinha Right; spin Hilbert spaces are finite-dimensional and one can, for the spin-$1/2$ Hilbert space for example, represent $S_z$ as a 2-by-2 matrix. There is no notion of momentum and position representations in these Hilbert spaces. $\endgroup$ – joshphysics Feb 4 '14 at 3:38
  • $\begingroup$ @RoopamSinha: In general, any linear operator on a separable Hilbert space admits a matrix representation. Regarding spin systems: it seems there is a bit of confusion, where you're mixing the bases of the states with the representation of the Hamiltonian. The state space in the spin system is a finite tensor product of finite-dimensional Hilbert spaces, which is ultimately a finite dimensional vector space. Thus any linear operator acting on it will admit a (finite) matrix representation. $\endgroup$ – user3657 Feb 4 '14 at 15:53
  • $\begingroup$ @joshphysics: Furthermore, every separable Hilbert space is isomorphic to $\ell^2(\mathbb{Z})$. Thus every QM Hamiltonian is unitarily equivalent to a Hamiltonian acting on $\ell^2(\mathbb{Z})$, where it admits a (rather easily constructed) Matrix representation. $\endgroup$ – user3657 Feb 4 '14 at 15:56

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