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1. What the question is asking (in plain language)

The student wants to know whether one can formulate a relativistic scalar field theory whose self‑interaction potential is a square‑well :

[ V(\phi)=\begin{cases} 0 & \text{if }-1<\phi<1,\[2pt] \infty & \text{otherwise}, \end{cases} \qquad\qquad (\;V(\phi)=\lim_{p\to\infty}\phi^{\,p}\;). ]

In other words the field value at every space‑time point is forced to stay inside the interval ((-1,1)).
The student tried to write down a lattice Hamiltonian, guessed a product of one‑particle box‑wave‑functions for the ground‑state functional, and then asked:

  • Is the theory well‑defined?
  • What does the exact ground‑state wave functional look like?
  • How do we construct excited states (e.g. a one‑particle state with momentum (k))?

Below we answer these points step by step, showing every algebraic and conceptual step.

2. Setting up the problem

2.1 Continuum Lagrangian

In (d) space‑time dimensions the action is

[ S[\phi]=\int d^dx\; \Bigl[\tfrac12(\partial_\mu\phi)(\partial^\mu\phi)-V(\phi)\Bigr], \qquad V(\phi)=\begin{cases} 0,&|\,\phi|<1,
\infty,&|\,\phi|\ge 1 . \end{cases} \tag{1} ]

Because (V) is infinite outside the interval, the functional integral (or the canonical Hilbert space) contains only those field configurations that satisfy

[ |\,\phi(x)|<1\quad\text{for every }x . \tag{2} ]

Thus the theory is nothing but a free scalar field with a hard bound on its amplitude.
In the language of sigma‑models the target space is the closed interval ([-1,1]) (or the open interval if we keep the walls strictly at (\pm1)).

2.2 Lattice regularisation

Put the theory on a hypercubic lattice of spacing (\varepsilon) (for simplicity a 1‑dimensional chain, the generalisation being straightforward).
At each site (n) we have a quantum‑mechanical coordinate (\phi_n).
The canonical Hamiltonian reads

[ H=\frac1N\sum_{n=1}^{N} \Bigl[ -\frac12\frac{\partial^{2}}{\partial\phi_n^{2}} +\frac12\frac{(\phi_n-\phi_{n-1})^{2}}{\varepsilon^{2}} +V(\phi_n) \Bigr], \qquad V(\phi_n)=\begin{cases} 0,&|\phi_n|<1,\[2pt] \infty,&\text{otherwise}. \end{cases} \tag{3} ]

The factor (1/N) in front of the sum is just a convenient overall normalisation; it does not affect the eigenfunctions.

3. Diagonalising the quadratic (kinetic) part

If we ignore the infinite walls for a moment, the Hamiltonian is a set of coupled harmonic oscillators.
Introduce discrete Fourier modes (normal‑mode coordinates)

[ \phi_n=\frac{1}{\sqrt N}\sum_{k}\;e^{ikn\varepsilon}\,q_k, \qquad k=\frac{2\pi}{N\varepsilon}m,\;m=0,\dots ,N-1 . \tag{4} ]

The gradient term becomes diagonal:

[ \frac12\sum_{n}\frac{(\phi_n-\phi_{n-1})^{2}}{\varepsilon^{2}} = \frac12\sum_{k}\omega_k^{2}\,q_k^{2}, \qquad \omega_k^{2}= \frac{4}{\varepsilon^{2}}\sin^{2}!\Bigl(\frac{k\varepsilon}{2}\Bigr). \tag{5} ]

The kinetic term (the second derivative with respect to (\phi_n)) also diagonalises:

[ -\frac12\sum_{n}\frac{\partial^{2}}{\partial\phi_n^{2}} =-\frac12\sum_{k}\frac{\partial^{2}}{\partial q_k^{2}} . \tag{6} ]

Hence, without the wall, the Hamiltonian splits into a sum of independent one‑dimensional particles:

[ H_{\text{free}}=\sum_{k}\Bigl[ -\frac12\frac{\partial^{2}}{\partial q_k^{2}} +\frac12\omega_k^{2}q_k^{2} \Bigr]. \tag{7} ]

The exact eigenfunctions are products of harmonic‑oscillator wavefunctions, (\displaystyle \Psi_{{n_k}}^{\text{free}}(q)=\prod_k\psi_{n_k}(q_k)).

4. Effect of the hard walls

The infinite potential (2) does not act on the normal‑mode variables separately; it is a constraint that couples all modes:

[ |\,\phi_n|<1\;\;\Longleftrightarrow\;\; \Bigl|\frac{1}{\sqrt N}\sum_{k}e^{ikn\varepsilon}q_k\Bigr|<1 \quad\forall n . \tag{8} ]

Thus the Hilbert space is the subspace of the free‑oscillator Hilbert space consisting of wavefunctions that vanish whenever any of the linear combinations (8) reaches (\pm1).

Consequences:

  • The Hamiltonian remains the sum (7); the only change is the domain of the wavefunctions.
  • The eigenfunctions are not products of the 1‑D box eigenfunctions (\sin(\pi\phi_n/2)) that the student guessed, because the coupling term mixes the sites.
  • The problem is equivalent to a free field with a compact target space; it belongs to the class of non‑linear sigma models with target ([-1,1]).

5. Ground‑state wave functional

5.1 General form

For a free scalar field (no walls) the exact ground‑state functional is the Gaussian

[ \boxed{ \Psi_{0}^{\text{free}}[\phi]=\mathcal N\; \exp!\Bigl[ -\frac12!\int!! d^{d-1}x\,d^{d-1}y\; \phi(\mathbf{x})\,K(\mathbf{x}-\mathbf{y})\,\phi(\mathbf{y}) \Bigr] } \tag{9} ]

with kernel (K) equal to the square root of the Laplacian (in momentum space (K(p)= p )).

5.2 Imposing the box constraint

Because the Hamiltonian does not change, the only effect of the walls is to restrict the support of the functional to the region

[ \mathcal R=\bigl{\phi(\mathbf{x})\;\big|\;|\phi(\mathbf{x})|<1\;\forall\mathbf{x}\bigr}. ]

Hence the exact ground state of the square‑well theory is

[ \boxed{ \Psi_{0}[\phi]= \begin{cases} \displaystyle \mathcal N\; \exp!\Bigl[ -\frac12!\int! d^{d-1}x\,d^{d-1}y\; \phi(\mathbf{x})\,K(\mathbf{x}-\mathbf{y})\,\phi(\mathbf{y}) \Bigr], & \phi\in\mathcal R,\[12pt] 0, & \text{otherwise}. \end{cases} } \tag{10} ]

In words: **the same Gaussian as for the free field, but truncated to the hyper‑cube ( \phi <1) at every point**.

The normalisation constant (\mathcal N) is fixed by

[ \int_{\mathcal R}!{\cal D}\phi\;|\Psi_{0}[\phi]|^{2}=1 . ]

Because the truncation removes only a tiny part of the infinite‑dimensional Gaussian measure (for a weakly fluctuating field), (\mathcal N) differs from the free‑field normalisation only by an exponentially small factor in the volume.

5.3 Why the product ansatz fails

The trial functional proposed in the question,

[ \Psi_{\text{trial}}[\phi]=\prod_{n}\cos!\Bigl(\tfrac{\pi}{2}\phi_n\Bigr), \tag{11} ]

has no dependence on the gradient term ((\phi_n-\phi_{n-1})^{2}).
Acting with the Hamiltonian (3) on (11) produces a non‑vanishing contribution from the gradient piece, because the derivative with respect to (\phi_n) brings down a factor (\tan(\frac{\pi}{2}\phi_n)) that is not cancelled by any term in the wavefunction.
Consequently (11) is not an eigenstate

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