In quantum mechanics we can express our wave function in terms of position space where the position operator, position eigenstates and wave function have the forms $ \hat{x} = x $, $|x \rangle = \delta(x-x') $ and $ \psi(x) $ respectively.

Now I have learned that a wave function can be thought of as existing independent to any basis we use which gives us the freedom to use any basis depending on the problem at hand, I.e. the momentum basis of momentum eigenstates to name one. This happens to be the Fourier transform of the position space wave function.

Well what if I decide to use energy eigenstates as my basis, is there such thing as an energy wave function $\psi(E)$ with the interpretation that $\psi(E)dE$ is the probability to find the system in the energy range $E$ to $E+ dE$? This space would have eigenstates of $|E \rangle = \delta(E-E')$. How would I transform from position to energy space?

  • 2
    $\begingroup$ Kind of. Unlike position and momentum, energy (Hamiltonian) often has discrete spectrum, so instead of energy "space" you get an energy scale, indexing the eigenstates. For the quantum oscillator they are the Hermite functions. When the Hamiltonian has continuous spectrum it will be more like $\psi(E)dE$, see projection-valued measures. $\endgroup$
    – Conifold
    Jul 12, 2017 at 0:19
  • $\begingroup$ To amplify this, in Fock space we add excitations of the QSHO, so |n> is an eigenstate of the number, hence energy operator. They constitute a complete orthonormal discrete set, alright. $\endgroup$ Jul 12, 2017 at 0:29
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    $\begingroup$ Yes the hermite functions are energy eigenstates but expressed in the position basis. What about a transform like the Fourier transform from position to momentum, but instead position to energy? $\endgroup$ Jul 12, 2017 at 0:30
  • $\begingroup$ To add up, the momentum space has a "dual" (not sure it's the appropriate name) which is its Fourier transform, namely the position space. Similarly, energy space has a "dual", time space. $\endgroup$
    – gingras.ol
    Jul 12, 2017 at 0:31
  • $\begingroup$ The "Fourier transform" into energy basis is just converting functions into the coefficients of their decompositions into Hermite functions, the inverse is summing up said decompositions. They are called generalized or abstract Fourier series. $\endgroup$
    – Conifold
    Jul 12, 2017 at 0:36

1 Answer 1


Yes, you can use eigenstates of the energy operator (the Hamiltonian) to express any state, but

i) The spectrum of the Hamiltonian can be discrete, so that instead of $|\psi\rangle = \int \psi(E)|E\rangle \, dE$, you have $|\psi\rangle = \sum \psi_E |E\rangle$. In fact, in introductory quantum mechanics when introducing the time independent Schrödinger equation, we use that any state can be decomposed as such.

ii) The spectrum of the Hamiltonian can be degenerate, so there is more than one state with the same energy. For example, for a free particle, the energy is $\mathbf p^2/2m$. It depends only on the magnitude of $\mathbf p$, not the direction. Therefore, a formula like $|\psi\rangle = \int \psi(E) |E\rangle$ isn't general enough. Which state with energy $E$ does $|E\rangle$ refer to? You need something like $|\psi\rangle = \int \psi(E, \hat p) |E, \hat p\rangle$ where $\hat p$ is a unit vector and signifies that the state has momentum in the $\hat p$ direction.

Since $E$ is, in this case, a function only of the magnitude of the momentum, we are in effect using the magnitude and direction of the momentum, so this is basically the same as momentum space. It is changing to spherical coordinates in momentum space followed by the change of variable $p \mapsto E = p^2/2m$. This is convenient for example when doing statistical physics because the Bose-Einstein and Fermi-Dirac distributions are functions of the energy, so it's convenient to use the energy as a variable.

iii) The momentum operator is always the same, so its eigenfunctions are always the same, and they are simple; they are plane waves.

On the other hand, the energy, the Hamiltonian operator, is different for each system -- a hydrogen-like atom, a harmonic oscillator, molecules, a crystal, a superconductor... In the general case, finding the eigenstates of the energy is hard and amounts to completely solving the dynamics of the system. You can see this from that if $$|\psi\rangle = \sum \tilde{c}_n(t) |E_n\rangle $$ and $\hat H|E_n\rangle = E_n |E_n\rangle$, then the Schrödinger equation gives $\tilde c_n(t) = c_n \exp(-iE_nt/\hbar)$ (the $|E_n\rangle$ should be time-independent). In the case of degeneracy, the $|E_n\rangle$ can be chosen orthogonal, so that $c_n = \langle E_n |\psi(t=0)\rangle = \int \psi_{E_n}^* \psi$. Hence you can get the time evolution for any initial state, that is, completely solve the dynamics.

iv) In relativistic theories, energy is the time component of the 4-momentum. Since the split between time and space is observer-dependent, in a relativistic theory, going over to momentum space one must also Fourier transform with respect to time. However, because it holds that $E^2 - p^2 = m^2$, only three components in momentum space are actually independent.

To summarize, yes, you can use eigenstates of the energy to expand the state (accounting for discrete and degenerate spectra), but finding the transition to that eigenbasis is equivalent to solving the Schrödinger equation. If you can do that, go ahead. If you cannot, you can still get somewhere with an energy eigenbasis if you only need general properties of the energy. This can be the case in thermodynamics, e.g. superconductors and Landau's theory of Fermi liquids.

  • $\begingroup$ Thanks for your explanation. How about negative energy contribution to the expansion? Are there eigenstates with negative values in the energy space representation of the wave function? $\endgroup$
    – N.S.
    Oct 15, 2020 at 0:55

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