Consider the harmonic oscillator hamiltonian $$\hat{H} = \frac{1}{2}(\hat{x}^2+\hat{p}^2)$$ where as usual $[\hat{x},\hat{p}] = i$ (with $\hbar=1$).

We could use the ladder ops $\hat{a}=\frac{1}{\sqrt{2}}(\hat{x}+i\hat{p})$ and $\hat{a}^{\dagger}=\frac{1}{\sqrt{2}}(\hat{x}-i\hat{p})$ to rewrite the Hamiltonian as $\hat{H} = \frac{1}{2}(\hat{a}^{\dagger}\hat{a}+1)$. Going through the usual argument, using the positivity of the norm and the commutation relations of $\hat{H}$ and $\hat{a}$, we conclude the existence of a vacuum state $\left| 0 \right>$ such that $\hat{a}\left|0\right> =0$. We choose $\left< 0 | 0\right> = 1$, and construct the Fock basis as $\left| n\right> = \frac{(\hat{a}^{\dagger})^n}{\sqrt{n!}}\left|0\right>$ such that $\left< n | m\right>=\delta_{nm}$. It is important to note that throughout this construction, we only assumed the positivity of the norm, and we did not specify concretely what the inner product on the Hilbert space is, in fact we derived the inner product expressed in terms of the Fock basis from other considerations.

Alternatively, if we had started in the Schodinger basis, namely, $\hat{x}\left|x\right> = x\left| x\right>$, $\left< x| x'\right> = \delta(x-x')$, and $\left<x\right|\hat{p}\left|x'\right> = -i\delta(x-x')\frac{\partial}{\partial x}$, we could solve for the eigenstates as expressed in the position basis. Note that in this case, the inner product is specified from the beginning by $\left< x| x'\right> = \delta(x-x')$, and not derived as in the Fock representation.

My question is, what guarantees that the Fock and Schrödinger Representations above are the same, i.e. they yield the same inner product? How do I know that in the Fock representation $\left< x| x'\right> = \delta(x-x')$ as well?

  • 2
    $\begingroup$ $\langle x | \hat{p} | x^\prime \rangle = -i \delta(x-x^\prime) \frac{\partial}{\partial x}$ does not make sense. $\endgroup$
    – Hyperon
    Jan 8 at 7:47
  • 2
    $\begingroup$ $\left<x\right|\hat{p}\left|x'\right> = -i\delta(x-x')\frac{\partial}{\partial x}$ is wrong, you got the derivative the wrong way (see Sakurai chap 1, eq 7.18 on the international edition). The correct expression is $\left<x\right|\hat{p}\left|x'\right> = -i\frac{\partial}{\partial x}\delta(x-x')$ $\endgroup$
    – peep
    Jan 8 at 8:05

1 Answer 1


The short answer is that these are two irreducible representations of the canonical commutation representations, so the Stone Von Neumann theorem guarantees the equivalence. In fact this correspondence is the key of the proof of the theorem.

More explicitly, the actual correspondence is given by the expression of the QHO energy eigenbasis in terms of position by the usual formula: $$ \langle x|n\rangle = \frac{1}{\sqrt{2^n n!}}\left(\frac{m\omega}{\pi\hbar}\right)^{1/4}e^{-m\omega x^2/2\hbar}H_n\left(\sqrt{\frac{m\omega }{\hbar}}x\right) $$

Btw, to avoid making mistakes with distribution, remember that they are a shorthand for linear forms, so: $$ \langle y|x\rangle = \delta(x-y) $$ really means for states $|\psi\rangle = \int dx \psi(x)|x\rangle$ and $|\phi\rangle = \int d^3x \phi(x)|x\rangle$: $$ \langle \phi|\psi\rangle = \int dx \phi(x)^*\psi(x) $$ Similarly: $$ \langle y|p|x\rangle = -i\hbar\delta'(x-y) $$ means: $$ \langle \phi|p|\psi\rangle = \int dx \phi(x)^*\left(-i\hbar\frac{d}{dx}\right)\psi(x) $$

Hope this helps.

  • $\begingroup$ Could you elaborate on the equivalence of the representations by constructing a unitary map between the two? $\endgroup$
    – Joeseph123
    Jan 8 at 13:02
  • 3
    $\begingroup$ @Joeseph123 Is not the formula for $\langle x|n\rangle$ the unitary map you want? $\sum_{n=0}^\infty \langle x|n\rangle \langle n|x'\rangle = \delta(x-x')$ after all. $\endgroup$
    – mike stone
    Jan 8 at 15:43
  • 4
    $\begingroup$ No. The $\langle x|n\rangle$ in @lpz 's answer are complete and orthonormal set with respect to the usual $L^2[{\mathbb R}]$ inner product. My formula is correct. No gaussians. $\endgroup$
    – mike stone
    Jan 8 at 19:12
  • 1
    $\begingroup$ Possibly related to this discussion. $\endgroup$ Jan 8 at 22:17
  • 1
    $\begingroup$ You have both $a^\dagger$ and $a$ -- but defining $a^\dagger$ requires an inner product. Conversely knowing what $a^\dagger$ is defines the inner product. $\endgroup$
    – mike stone
    Jan 10 at 13:48

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