Schrödinger Wave Functionals

The Schrödinger representation describes a quantum field state by its complex amplitude on all field configurations at one time. With a finite real-mode regulator, that “wave functional” is an ordinary wave function of finitely many coordinates: the field multiplies it, the canonical momentum differentiates it, and the free Hamiltonian is a sum of harmonic-oscillator Hamiltonians. The massive free vacuum is therefore a normalized Gaussian whose exponent contains the positive square root of the spatial Klein–Gordon operator. Its inverse kernel is twice the equal-time covariance.

The construction below uses a real scalar of mass $m>0$ in a periodic spatial box, retaining $M<\infty$ independent real modes. It treats configuration and momentum representations, the Gaussian vacuum, oscillator excitations, and finite boundary wave functions. Continuum functional notation is introduced only as shorthand for that regulated system. No flat infinite-dimensional Lebesgue measure, interacting variational algorithm, or regulator-removal theorem is claimed.

Required background. Canonical Quantization: Algebra, Representation, and State supplies the canonical relations and the distinction between a state and its representation. Quantizing the Real Scalar Field supplies the normalized free Hamiltonian, vacuum choice, and oscillator-mode decomposition used below.

A field state becomes a function of real mode amplitudes

Let $\{f_\alpha(\mathbf x)\}_{\alpha=1}^{M}$ be real orthonormal functions in the periodic box. Choose them to diagonalize the regulated positive spatial operator:

$$\begin{aligned} \mathcal K_\Lambda f_\alpha &= \omega_\alpha^2 f_\alpha, \\ \int_V \mathrm d^{d-1}\mathbf x\, f_\alpha(\mathbf x)f_\beta(\mathbf x) &= \delta_{\alpha\beta}, \\ \omega_\alpha &>0 . \end{aligned}$$

For the free scalar, $\mathcal K_\Lambda$ is the finite-mode restriction of $-\nabla^2+m^2$. Expand the Schrödinger-picture operators as

$$\begin{aligned} \widehat\phi_\Lambda(\mathbf x) &= \sum_{\alpha=1}^{M} \widehat q_\alpha f_\alpha(\mathbf x), \\ \widehat\pi_\Lambda(\mathbf x) &= \sum_{\alpha=1}^{M} \widehat p_\alpha f_\alpha(\mathbf x). \end{aligned}$$

The mode operators obey

$$[\widehat q_\alpha,\widehat p_\beta] = i\delta_{\alpha\beta}, \qquad [\widehat q_\alpha,\widehat q_\beta] = [\widehat p_\alpha,\widehat p_\beta] = 0 .$$

For a periodic Fourier regulator, a real basis contains the constant mode and one cosine and one sine mode for each retained nonzero pair $\{\mathbf k,-\mathbf k\}$. Equivalently, complex Fourier coefficients obey $\varphi_{-\mathbf k}=\varphi_{\mathbf k}^*$. Treating both members of every pair as independent complex coordinates would double the number of real degrees of freedom.

Let $|\mathbf q\rangle$ be a generalized simultaneous eigenket, with $\mathbf q=(q_1,\ldots,q_M)\in\mathbb R^M$, such that

$$\widehat q_\alpha|\mathbf q\rangle = q_\alpha|\mathbf q\rangle \qquad(\alpha=1,\ldots,M).$$

The regulated wave functional is the ordinary coordinate wave function

$$\Psi_\Lambda[\varphi,t] \equiv \Psi_\Lambda(\mathbf q,t) = \langle\mathbf q|\Psi(t)\rangle,$$

where the argument field is

$$\varphi_\Lambda(\mathbf x) = \sum_{\alpha=1}^{M} q_\alpha f_\alpha(\mathbf x).$$

With delta-normalized configuration kets, the state norm is

$$\|\Psi\|^2 = \int_{\mathbb R^M} \mathrm d^M q\, |\Psi_\Lambda(\mathbf q,t)|^2 .$$

The Schwartz space $\mathcal S(\mathbb R^M)$ is a convenient common invariant core for the coordinate, momentum, and oscillator Hamiltonian operators. Configuration eigenkets themselves are distributions, not normalizable Hilbert-space vectors.

The finite-dimensional calculation proceeds as follows.

• Input. Supply a finite set of independent real canonical pairs, a positive quadratic stiffness matrix, and the state or boundary condition to represent.
• Procedure. Diagonalize the stiffness matrix, represent momenta by derivatives, solve the resulting oscillator Schrödinger equations, and transform back to the chosen real field basis.
• Output. Obtain a normalized finite-dimensional wave function, its energy and covariance, and—when desired—its momentum-space transform and oscillator excitations.
• Cost. Dense diagonalization of an $M\times M$ matrix costs $O(M^3)$ operations and $O(M^2)$ storage. Translation invariance makes the Fourier basis diagonal from the start.
• Validation. Check the canonical commutator, normalization, the Schrödinger equation, the covariance, and agreement with the regulated Fock-mode spectrum.
• Stop rule. A zero eigenvalue, removal of the regulator, a nonquadratic interaction, or a demand for a rigorous infinite-dimensional measure requires additional analysis.

The functional Schrödinger equation is a finite oscillator equation

In the configuration representation,

$$\begin{aligned} \widehat q_\alpha\Psi_\Lambda &= q_\alpha\Psi_\Lambda, \\ \widehat p_\alpha\Psi_\Lambda &= -i\frac{\partial\Psi_\Lambda}{\partial q_\alpha}. \end{aligned}$$

The regulated functional derivative is only an abbreviation for these ordinary derivatives:

$$\frac{\delta_\Lambda}{\delta\varphi(\mathbf x)} = \sum_{\alpha=1}^{M} f_\alpha(\mathbf x) \frac{\partial}{\partial q_\alpha}.$$

It differentiates the projected configuration according to

$$\frac{\delta_\Lambda \varphi_\Lambda(\mathbf x)} {\delta\varphi(\mathbf y)} = P_\Lambda(\mathbf x,\mathbf y),$$

where

$$P_\Lambda(\mathbf x,\mathbf y) = \sum_{\alpha=1}^{M} f_\alpha(\mathbf x)f_\alpha(\mathbf y)$$

is the finite-mode projector kernel, not yet the continuum delta distribution. Consequently,

$$\widehat\pi_\Lambda(\mathbf x) \Psi_\Lambda = -i \frac{\delta_\Lambda\Psi_\Lambda} {\delta\varphi(\mathbf x)},$$

and the represented equal-time commutator is

$$[\widehat\phi_\Lambda(\mathbf x), \widehat\pi_\Lambda(\mathbf y)] = iP_\Lambda(\mathbf x,\mathbf y).$$

The regulated free Hamiltonian is

$$\widehat H_\Lambda = \frac12 \sum_{\alpha=1}^{M} \left( \widehat p_\alpha^2 + \omega_\alpha^2\widehat q_\alpha^2 \right).$$

It acts on wave functions as

$$\widehat H_\Lambda = \frac12 \sum_{\alpha=1}^{M} \left( -\frac{\partial^2}{\partial q_\alpha^2} + \omega_\alpha^2q_\alpha^2 \right),$$

so the functional Schrödinger equation is the finite partial differential equation

$$i\frac{\partial}{\partial t} \Psi_\Lambda(\mathbf q,t) = \widehat H_\Lambda \Psi_\Lambda(\mathbf q,t).$$

No new quantization postulate has appeared: this is a representation of the same finite canonical algebra and Hamiltonian used in the oscillator and Fock pictures.

For each mode, define

$$\widehat a_\alpha = \frac{1}{\sqrt{2\omega_\alpha}} \left( \omega_\alpha\widehat q_\alpha + i\widehat p_\alpha \right).$$

In the configuration representation, the vacuum condition becomes, for every $\alpha=1,\ldots,M$,

$$\left( \frac{\partial}{\partial q_\alpha} + \omega_\alpha q_\alpha \right) \Psi_{0,\Lambda} = 0.$$

The positive-frequency annihilators—and not the canonical commutator alone—supply the state condition that selects the Gaussian solution.

The vacuum contains the positive square-root kernel

Solving the first-order equation for one mode and normalizing it in $L^2(\mathbb R,\mathrm d q_\alpha)$ gives

$$\psi_{0,\alpha}(q_\alpha) = \left( \frac{\omega_\alpha}{\pi} \right)^{1/4} \exp\left( -\frac12\omega_\alpha q_\alpha^2 \right).$$

The full regulated vacuum is the unambiguous product of these normalized single-mode factors:

$$\Psi_{0,\Lambda}(\mathbf q) = \prod_{\alpha=1}^{M} \psi_{0,\alpha}(q_\alpha).$$

Let $\Omega_\Lambda$ be the positive square root of the stiffness matrix:

$$\Omega_\Lambda^2 = \mathcal K_\Lambda, \qquad \Omega_\Lambda>0 .$$

In the diagonal real-mode basis,

$$(\Omega_\Lambda)_{\alpha\beta} = \omega_\alpha\delta_{\alpha\beta}.$$

The same vacuum can therefore be written

$$\Psi_{0,\Lambda}(\mathbf q) = \mathcal N_\Lambda \exp\left( -\frac12 \mathbf q^{\mathsf T} \Omega_\Lambda \mathbf q \right),$$

with

$$\mathcal N_\Lambda = \left( \frac{\det\Omega_\Lambda}{\pi^M} \right)^{1/4}.$$

The position-space kernel on the retained subspace is

$$\Omega_\Lambda(\mathbf x,\mathbf y) = \sum_{\alpha=1}^{M} \omega_\alpha f_\alpha(\mathbf x)f_\alpha(\mathbf y).$$

Thus the exponent has the equivalent form

$$\begin{gathered} \mathbf q^{\mathsf T}\Omega_\Lambda\mathbf q \\ = \int_V \mathrm d^{d-1}\mathbf x \int_V \mathrm d^{d-1}\mathbf y\; \\[-0.25em] {}\quad\times\varphi_\Lambda(\mathbf x) \Omega_\Lambda(\mathbf x,\mathbf y) \varphi_\Lambda(\mathbf y). \end{gathered}$$

For a translation-invariant regulator, the Fourier symbol of this kernel is $\omega_{\mathbf k}=\sqrt{\mathbf k^2+m^2}$. The often-written continuum formula

$$\Omega = \sqrt{-\nabla^2+m^2}$$

means the positive spectral square root. Although $-\nabla^2+m^2$ is a local differential operator, its square root is nonlocal in position space. Introduce the formal quadratic form

$$\begin{gathered} \mathcal Q[\varphi] \equiv \int\mathrm d^{d-1}\mathbf x \int\mathrm d^{d-1}\mathbf y\; \\[-0.25em] {}\quad\times \varphi(\mathbf x) \Omega(\mathbf x,\mathbf y) \varphi(\mathbf y). \end{gathered}$$

Then

$$\Psi_0[\varphi] \propto \exp\left(-\frac12\mathcal Q[\varphi]\right)$$

is therefore a compact continuum mnemonic for the finite-mode Gaussian, not a definition using an infinite product of ordinary Lebesgue measures. The mode-by-mode field-eigenstate construction appears in Srednicki 2006 manuscript, § 8, problem 8.8, p. 70. The variational-derivative representation, vacuum condition, Gaussian ansatz, and square-root kernel are developed in Weinberg 1995, vol. I, § 9.2, pp. 385–388.

Energy and covariance reproduce the Fock vacuum

One oscillator supplies three direct checks. For $\omega>0$,

$$\psi_0(q) = \left( \frac{\omega}{\pi} \right)^{1/4} e^{-\omega q^2/2}.$$

Its normalization is

$$\int_{-\infty}^{\infty} \mathrm dq\, |\psi_0(q)|^2 = 1.$$

Direct differentiation gives

$$\frac12 \left( -\frac{\mathrm d^2}{\mathrm dq^2} + \omega^2q^2 \right) \psi_0(q) = \frac{\omega}{2}\psi_0(q),$$

and the Gaussian variance is

$$\int_{-\infty}^{\infty} \mathrm dq\, q^2|\psi_0(q)|^2 = \frac{1}{2\omega}.$$

For all retained modes,

$$E_{0,\Lambda} = \frac12 \operatorname{tr}\Omega_\Lambda = \frac12 \sum_{\alpha=1}^{M}\omega_\alpha,$$

while

$$\langle0| \widehat q_\alpha\widehat q_\beta |0\rangle = \frac12 (\Omega_\Lambda^{-1})_{\alpha\beta}.$$

The equal-time field covariance is consequently

$$\begin{aligned} \langle0| \widehat\phi_\Lambda(\mathbf x) \widehat\phi_\Lambda(\mathbf y) |0\rangle = \frac12 \Omega_\Lambda^{-1} (\mathbf x,\mathbf y). \end{aligned}$$

This is exactly the equal-time limit of the regulated positive-frequency two-point function. The energy is the same zero-point sum and the covariance is the same $1/(2\omega_\alpha)$ mode variance found in the Fock representation. Removing the scalar energy by normal ordering is a separate reference-vacuum prescription, treated on Normal Ordering and Vacuum Terms.

There is also a basis-independent check. Insert the Gaussian ansatz

$$\Psi_A(\mathbf q) = \mathcal N_A \exp\left( -\frac12\mathbf q^{\mathsf T}A\mathbf q \right),$$

where $A=A^{\mathsf T}>0$, into a quadratic Hamiltonian

$$\widehat H = \frac12 \left( -\nabla_{\mathbf q}^2 + \mathbf q^{\mathsf T} \mathcal K \mathbf q \right).$$

Matching the quadratic and constant terms yields

$$A^2 = \mathcal K, \qquad E_0 = \frac12\operatorname{tr}A.$$

Normalizability selects the positive solution $A=\mathcal K^{1/2}$, independently reproducing the mode diagonalization.

For a concrete coupled two-mode example, take

$$\mathcal K = \begin{pmatrix} 3&1\\ 1&2 \end{pmatrix}.$$

It is positive definite and has eigenvalues

$$\lambda_\pm = \frac{5\pm\sqrt5}{2}.$$

The two frequencies and ground energy are

$$\omega_\pm = \sqrt{\lambda_\pm}, \qquad E_0 = \frac12 \left( \omega_++\omega_- \right).$$

Using $A=\mathcal K^{1/2}$ in the Gaussian gives covariance $\frac12\mathcal K^{-1/2}$. Diagonalizing first gives two independent oscillators with the same frequencies, energy, and covariance. Exact symbolic differentiation and Gaussian integration reproduce the normalization, Schrödinger equation, and $1/(2\omega)$ variance; this checks only the finite algebra, not a continuum measure or an interactive calculation.

Momentum wave functions and excited states use the same oscillators

The unitary finite-dimensional Fourier kernel is

$$F_M(\mathbf p,\mathbf q) = \frac{e^{-i\mathbf p\cdot\mathbf q}}{(2\pi)^{M/2}}.$$

It defines the momentum representation by

$$\widetilde\Psi_\Lambda(\mathbf p,t) = \int_{\mathbb R^M} \mathrm d^M q\, F_M(\mathbf p,\mathbf q) \Psi_\Lambda(\mathbf q,t).$$

There,

$$\begin{aligned} \widehat p_\alpha\widetilde\Psi_\Lambda &= p_\alpha\widetilde\Psi_\Lambda, \\ \widehat q_\alpha\widetilde\Psi_\Lambda &= i\frac{\partial\widetilde\Psi_\Lambda} {\partial p_\alpha}. \end{aligned}$$

Fourier transformation sends the vacuum to

$$\widetilde\Psi_{0,\Lambda}(\mathbf p) = \widetilde{\mathcal N}_\Lambda \exp\left( -\frac12 \mathbf p^{\mathsf T} \Omega_\Lambda^{-1} \mathbf p \right),$$

where

$$\widetilde{\mathcal N}_\Lambda = \left( \frac{1} {\pi^M\det\Omega_\Lambda} \right)^{1/4}.$$

Thus narrow configuration fluctuations at large $\omega_\alpha$ correspond to broad momentum fluctuations, and vice versa.

The normalized occupation-basis wave functions are

$$\begin{aligned} \Psi_{\{n_\alpha\}}(\mathbf q) = \prod_{\alpha=1}^{M} &\frac{ H_{n_\alpha}(\sqrt{\omega_\alpha}q_\alpha) }{ \sqrt{2^{n_\alpha}n_\alpha!} } \\ &\times \Psi_{0,\Lambda}(\mathbf q), \end{aligned}$$

with energies

$$E_{\{n_\alpha\}} = E_{0,\Lambda} + \sum_{\alpha=1}^{M} n_\alpha\omega_\alpha.$$

The Hermite polynomial factors are the configuration-representation images of applying Fock creation operators. Their products form the usual oscillator basis of $L^2(\mathbb R^M)$. The Fock and wave-functional pictures are therefore two representations of the same matched regulated system, not two different free theories.

Boundary wave functions select the state

A stationary vacuum evolves as

$$\Psi_{0,\Lambda}(\mathbf q,t) = e^{-iE_{0,\Lambda}t} \Psi_{0,\Lambda}(\mathbf q).$$

More generally, let

$$K_\Lambda( \mathbf q_f,t_f; \mathbf q_i,t_i) = \langle\mathbf q_f| e^{-i\widehat H_\Lambda(t_f-t_i)} |\mathbf q_i\rangle$$

be the finite configuration-space kernel. An amplitude between arbitrary boundary states is $\mathcal A_{fi}=\langle\Psi_f| e^{-i\widehat H_\Lambda(t_f-t_i)}|\Psi_i\rangle$. For brevity, set

$$\begin{gathered} K_\Lambda^{fi}(\mathbf q_f,\mathbf q_i) \\ \equiv K_\Lambda(\mathbf q_f,t_f;\mathbf q_i,t_i). \end{gathered}$$

Define the corresponding integrand by

$$\mathcal I_{fi}(\mathbf q_f,\mathbf q_i) = \Psi_f^*(\mathbf q_f) K_\Lambda^{fi}(\mathbf q_f,\mathbf q_i) \Psi_i(\mathbf q_i).$$

Then

$$\mathcal A_{fi} = \int\mathrm d^M q_f \int\mathrm d^M q_i\; \mathcal I_{fi}(\mathbf q_f,\mathbf q_i).$$

The bulk Hamiltonian determines propagation, but its differential equation admits the vacuum, excited states, and arbitrary superpositions. The boundary wave functions specify which state is prepared. For the vacuum, their positive square-root kernel records both the positive-frequency choice and the equal-time covariance. In a gapped finite system, long Euclidean evolution can project suitable boundary data onto the ground state; that projection and its normalization are additional boundary prescriptions, not consequences of writing the Lorentzian bulk action alone.

The finite transition kernel is developed from time slicing on Time Slicing and Transition Amplitudes. The role of initial and final state factors continues on Boundaries and State Preparation. The exact conditions under which the canonical and functional calculations agree belong to the Canonical–Functional Crosswalk for Regulated Systems.

Where the finite construction stops

Continuum normalization. As $M\to\infty$, the determinant in $\mathcal N_\Lambda$ and the product $\prod_\alpha\mathrm dq_\alpha$ do not automatically define a flat measure on a space of fields. Start the functional description from a regulator on Regulated Bosonic Field Integrals. Gaussian Euclidean measures provide a related rigorous construction, but they are not obtained by declaring an infinite-dimensional Lebesgue measure; see Gaussian Euclidean Fields as Measures.

Zero modes. The derivation requires every $\omega_\alpha>0$. For a massless scalar in a periodic box, the spatially constant mode has $\omega_{\mathbf0}=0$, so its proposed “Gaussian” is constant and not normalizable on $\mathbb R$. The infrared alternatives and order of limits are treated on Massless Scalars, Zero Modes, and Infrared Limits.

Interactions. Adding a nonquadratic potential turns the Schrödinger equation into a coupled many-variable problem. A Gaussian trial functional may become a variational ansatz, but the free square-root kernel no longer solves the exact equation. Systematic field-theory ansätze and their error criteria continue on Variational Principles and Field-Theory Ansätze.

Representation changes. A different positive-frequency split changes the annihilators and hence the vacuum Gaussian. In systems with infinitely many degrees of freedom, formally related canonical variables need not give unitarily equivalent Hilbert-space representations. The finite calculation does not prove continuum equivalence.

Common pitfalls

Treating all complex Fourier coefficients as independent. A real field obeys $\varphi_{-\mathbf k}=\varphi_{\mathbf k}^*$. Use independent real cosine and sine amplitudes, or choose one representative of each nonzero momentum pair with its two real components.

Replacing the finite projector by a delta too early. At finite regulator, the derivative of one configuration value with respect to another is $P_\Lambda(\mathbf x,\mathbf y)$. It approaches a delta distribution only under a stated regulator limit.

Using the stiffness matrix instead of its square root. The Hamiltonian contains $\mathcal K=\Omega^2$, while the vacuum exponent contains $\Omega=\mathcal K^{1/2}$. Confusing them gives the wrong energy and equal-time covariance.

Saying that the bulk action selects the vacuum. The same Hamiltonian evolves every state. The ground-state condition, asymptotic prescription, or boundary wave function supplies the state-selection data.

Calling the formal continuum product a measure. The finite product $\mathrm d^M q$ is ordinary Lebesgue measure. Its infinite-dimensional analogue requires a different mathematical construction.

Check your understanding

1. Retrieval. Define $\Psi_\Lambda[\varphi,t]$ and state how $\widehat\phi_\Lambda(\mathbf x)$ and $\widehat\pi_\Lambda(\mathbf x)$ act on it.
2. Distinction. Separate the field configuration $\varphi_\Lambda$, the quantum state $|\Psi\rangle$, its coordinate wave function $\Psi_\Lambda$, and a functional-integral history.
3. Derivation. Starting from one oscillator Hamiltonian, derive the normalized ground-state Gaussian, energy $\omega/2$, and variance $1/(2\omega)$.
4. Failure diagnosis. Explain both errors in integrating independently over $\varphi_{\mathbf k}$ and $\varphi_{-\mathbf k}$ for a real field and in writing a normalizable vacuum factor for a mode with $\omega=0$.
5. Transfer. For a positive matrix $\mathcal K$, insert a Gaussian with kernel $A$ into the Schrödinger equation and derive the equation that fixes $A$ and $E_0$.
6. Handoff. Decide which continuation is needed for a time-sliced transition amplitude, a rigorous Gaussian measure, an interacting variational ansatz, and a massless periodic zero mode.

Answers and repair routes

1. $\Psi_\Lambda[\varphi,t]=\langle\mathbf q|\Psi(t)\rangle$, with $\varphi_\Lambda=\sum_\alpha q_\alpha f_\alpha$. The field multiplies by $\varphi_\Lambda(\mathbf x)$ and the momentum acts as $-i\delta_\Lambda/\delta\varphi(\mathbf x)$. Repair the canonical relation at Canonical Quantization: Algebra, Representation, and State.
2. $\varphi_\Lambda$ is one classical argument of the wave function; $|\Psi\rangle$ is the abstract state; $\Psi_\Lambda$ is that state’s representative in the configuration basis; and a history assigns a configuration at every time in a regulated transition calculation. Repair the representation distinction at Canonical Quantization: Algebra, Representation, and State and the history distinction at Time Slicing and Transition Amplitudes.
3. The equation $(\partial_q+\omega q)\psi_0=0$ gives $\psi_0\propto e^{-\omega q^2/2}$; Gaussian normalization supplies $(\omega/\pi)^{1/4}$. Two derivatives give $H\psi_0=(\omega/2)\psi_0$, and the normalized Gaussian integral gives $\langle q^2\rangle=1/(2\omega)$. Repair the oscillator decomposition at Quantizing the Real Scalar Field.
4. Reality relates the two complex Fourier coefficients, so treating them as independent doubles the real coordinates. At $\omega=0$, the exponential has no restoring quadratic term and is constant along that coordinate, so it is not square-integrable. Repair the infrared case at Massless Scalars, Zero Modes, and Infrared Limits.
5. Differentiating $\exp(-\mathbf q^{\mathsf T}A\mathbf q/2)$ produces a quadratic term proportional to $A^2$. Matching it to the potential gives $A^2=\mathcal K$ and $E_0=\frac12\operatorname{tr}A$; normalizability selects $A=\mathcal K^{1/2}>0$. Repair the finite-mode Hamiltonian at Quantizing the Real Scalar Field.
6. Use Time Slicing and Transition Amplitudes for the transition kernel; Gaussian Euclidean Fields as Measures for the rigorous measure problem; Variational Principles and Field-Theory Ansätze for interacting trial states; and Massless Scalars, Zero Modes, and Infrared Limits for the periodic zero mode.

References

• Srednicki, Mark. Quantum Field Theory. Author-hosted manuscript, University of California, Santa Barbara, 2006. Author’s manuscript page.
• Weinberg, Steven. The Quantum Theory of Fields. Volume I: Foundations. Cambridge University Press, 1995. doi:10.1017/CBO9781139644167.