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A real centered Gaussian wave packet with initial position variance $1/(2\alpha )$ and momentum variance ${\hslash }^2\alpha /2$ evolves under a quantum harmonic oscillator byThe covariance matrix obeys $\Sigma (t)=M\Sigma (0)M^{\mathsf{T}}$ and $\det \Sigma ={\hslash }^2/4$. Thus the Robertson-Schrödinger uncertainty relation stays saturated, although the position-momentum variance product can exceed ${\hslash }^2/4$ because of nonzero quantum covariance. The widths are stationary only for $\alpha =m\omega /\hslash$; otherwise they breathe with frequency $2\omega$.

A chirped Gaussian wave packet adds a quadratic quantum phase to the zero-mean Gaussian wave packet already evaluated above. Here $\sigma >0$ and $\kappa$ is a real dimensionless parameter. Its probability density function $|{\psi }_{\sigma ,\kappa }(x){|}^2$ is unchanged for every $\kappa$, so $⟨x⟩=0$ and $(\Delta x{)}^2={\sigma }^2$. Nevertheless, applying the momentum operator givesThe sign of the quantum covariance follows directly, without assigning a classical trajectory to the wavefunction:Here $r$ is a normalized symmetrized quantum covariance, analogous to a correlation coefficient. Positive $\kappa$ gives positive $C_{xp}$, and reversing $\kappa$ reverses $C_{xp}$ while leaving both variances unchanged. In the negative-exponent Fourier transform convention used above, the momentum density isConsequently the quantum uncertainty product and the covariance matrix determinant tell different stories:Every member of this pure state family saturates the Robertson-Schrödinger uncertainty relation. Only $\kappa =0$ saturates the Heisenberg uncertainty relation for this fixed pair of position and momentum axes. Thus being a Gaussian wave packet alone does not imply a minimum $\Delta x\Delta p$.

For a concrete comparison, choose a reference length $L$, put $X=x/L$, $P=pL/\hslash$, and set $\sigma =L$. The choices $\kappa =0,+2,-2$ all have $(\Delta X{)}^2=1$, but give $(\Delta P{)}^2=1/4,5/4,5/4$ and $C_{XP}=0,+1,-1$, respectively. The two nonzero cases have $r=\pm 2/\sqrt{5}$ and product $\sqrt{5}/2$; all three covariance matrices have determinant $1/4$.

The phase space curves in the figure satisfy $z^T{\Sigma }^{-1}z=1$ in the dimensionless coordinates, where $z=(X,P{)}^T$. They display second moments, not observed simultaneous position and momentum pairs. Their common area is $\pi \sqrt{\det \Sigma }=\pi /2$, although their coordinate projections differ.

This distinction is useful for predicting focusing. Under free particle evolution with mass $m$, $x(t)=x(0)+tp(0)/m$, soA negative chirp initially narrows the position distribution; positive chirp initially broadens it. For $\kappa <0$, the minimum occurs at $t_{\ast }=-2m{\sigma }^2\kappa /[\hslash (1+{\kappa }^2)]>0$ and has variance ${\sigma }^2/(1+{\kappa }^2)$. At that time the quantum covariance vanishes, and the unchanged momentum spread gives $\Delta x(t_{\ast })\Delta p=\hslash /2$. The extra initial product was a usable position-momentum correlation, not extra measurement backaction.

Equality in the Heisenberg uncertainty relation requires the centred vectors $x^{\prime }\psi$ and $p^{\prime }\psi$ to be imaginary scalar multiples, where $x^{\prime }=x-x_{0}I$, $p^{\prime }=p-p_{0}I$ and $x_0,p_0$ are the expectation values. Thus, for some $s>0$,In the position representation, using the momentum operator turns this into a first-order differential equation, whose normalizable solutions areHence the quantum states saturating the Heisenberg uncertainty relation are these Gaussian wave packets, with arbitrary position and momentum means and global phase. Their width is ${\sigma }^2=\hslash /(2s)$. A Gaussian wave packet with an additional quadratic quantum phase need not saturate this product bound.

Let the initial Gaussian wave packet beFourier transforming, multiplying each momentum amplitude by $e^{-i{p}^{2}t/(2m\hslash )}$ and evaluating the Gaussian integral givesIts probability density is a normalized Gaussian with center $x_0+p_{0}t/m$ and width ${\sigma }_x={\sigma }_0\sqrt{1+{\eta }^2}$; the momentum width remains ${\sigma }_p=\hslash /(2{\sigma }_0)$. Its symmetrized position-momentum covariance is $C_{xp}=t{\sigma }_p^2/m$, soThus free spreading preserves equality in the Robertson-Schrödinger uncertainty relation while producing a nonzero quadratic position-dependent phase and a product ${\sigma }_x{\sigma }_p$ larger than $\hslash /2$. More generally, ${\sigma }_x(t{)}^2={\sigma }_x(0{)}^2+2t{C}_{xp}(0)/m+t^2{\sigma }_p(0{)}^2/m^2$, so an initially correlated Gaussian can first contract.

Under a quantum harmonic oscillator Hamiltonian, the Heisenberg equations of motion giveSubtract the expectation values, square the first equation, and use the symmetrized covariance $C_0=⟨\{x-⟨x⟩,p-⟨p⟩\}⟩/2$. This yieldsThese moment equations hold for any state with finite second moments; a quadratic Hamiltonian additionally preserves Gaussian wave packets. A coherent state has ${\sigma }_x(0{)}^2=\hslash /(2m\omega )$, ${\sigma }_p(0{)}^2=m\hslash \omega /2$, and $C_0=0$, so its width stays fixed while its center follows the classical orbit. A squeezed coherent state generally has a width that oscillates at $2\omega$.

For the Hamiltonian $H=(p^2-x^2)/2$, a Gaussian wave packet $A{e}^{-B{x}^2}$ satisfies $\dot{A}=-i\hslash AB$ and $\dot{B}=-i/(2\hslash )-2i\hslash B^2$. The displayed solution is normalizable for real $\varphi$ exactly when $\sin (2\varphi )>0$; its second moments of a complex Gaussian wave packet areBoth grow as $\hslash e^{2t}/[4\sin (2\varphi )]$, while their difference and hence the mean energy remain constant.

Use the quantum harmonic oscillator in units with reduced Planck constant $\hslash =1$, mass $m=1$, and angular frequency $\omega =1$. Its Hamiltonian operator and trial Gaussian wave packet areThis is the same Gaussian wave packet used for the Heisenberg uncertainty relation: $|{\psi }_{\sigma }{|}^2$ is a normal distribution with mean zero and variance ${\sigma }^2$. The Gaussian integral gives wavefunction normalization and $⟨x^2⟩={\sigma }^2$. These smooth, rapidly decaying trials lie in the one-dimensional harmonic oscillator form domain and even in the operator domain.

Since ${\psi }_{\sigma }^{\prime }=-x{\psi }_{\sigma }/(2{\sigma }^2)$, integration by parts with vanishing boundary terms gives the kinetic energy and potential energy:Narrow trials cost kinetic energy, while broad trials cost potential energy. The exact differenceshows that the unique optimal width is ${\sigma }^2=1/2$, with $E=1/2$. At $\sigma =1/2$ and $\sigma =1$, the trial energies are both $5/8=0.625$; the kinetic energy and potential energy contributions exchange places.

Minimizing within this trial family alone establishes $E_0\le 1/2$. To establish exactness, use the Heisenberg uncertainty relation to obtain $E_{\psi }\ge 1/2$ for every admissible quantum state, as in its existing oscillator example. Together the bounds give $E_0=1/2$. Direct substitution also yields $H{\psi }_{1/\sqrt{2}}={\psi }_{1/\sqrt{2}}/2$, agreeing with the Gaussian eigenstate for a quadratic potential. This is a useful distinction: the variational method supplies the upper bound, while the Heisenberg uncertainty relation supplies the complementary lower bound.

The ground-state energy $E_0$ is the energy eigenvalue attained by a ground state. For a self-adjoint operator $H$ bounded below, distinguish this attained value from the always-defined spectral bottom $E_{\ast }=\inf \sigma (H)$. When a ground state exists, $E_0=E_{\ast }$. Some usage calls $E_{\ast }$ the ground-state energy even when it is unattained; here an unattained bottom is explicitly identified as an infimum. The quantum variational principle statesApproaching this infimum with trial wavefunctions does not ensure that a minimizing wavefunction exists. ^[ref].

For the whole-line free particle with mass $m>0$,Its spectrum is continuous and contains no energy eigenvalues represented by normalizable wavefunctions. The Fourier transform makes $H$ multiplication by ${\hslash }^2{k}^2/(2m)$; a fixed energy restricts $k$ to a set of zero length, on which a nonzero square-integrable function cannot be supported. ^[ref]. An explicit family of normalized Gaussian wave packets demonstrates how the bottom is approached:The formula follows from ${\psi }_s^{\prime }=-x{\psi }_s/(2s^2)$, integration by parts, and $⟨x^2{⟩}_s=s^2$. Every finite width gives positive kinetic energy. At zero energy, $H\psi =0$ would require an affine function on the whole line, which is never a nonzero normalizable wavefunction. Spreading a Gaussian wave packet indefinitely produces an unattained infimum, rather than a zero-energy ground state.

An additive choice of energy origin changes the numerical ground-state energy without changing the ground-state subspace. For $H^{\prime }=H+CI$, every energy eigenvalue becomes $E^{\prime }=E+C$, while all differences $E_j-E_i$ stay fixed. Time evolution gains only the global phase $e^{-iCt/\hslash }$. This makes excitation costs and spectroscopic differences more intrinsic than the isolated number $E_0$.

The Heisenberg uncertainty relation constrains the standard deviations of position and momentum in one preparation:On the real line, the position operator $x$ acts by multiplication, and the momentum operator is $p=-i\hslash d/dx$, with $\hslash$ the reduced Planck constant. Their canonical commutation relation $[x,p]=i\hslash I$ and the Robertson uncertainty principle give the bound on normalized quantum states satisfying the moment and product-domain assumptions above; Schwartz functions are a convenient sufficient class. Thus making a preparation more sharply localized in position forces a wider momentum distribution, even for ideal measurement in quantum mechanics.

An explicit saturating example is the normalized Gaussian wave packet, for $\sigma >0$,The wavefunction is a probability amplitude; its squared magnitude is the probability density function, by the Born rule. That density is a normal distribution of mean zero and standard deviation $\Delta x=\sigma$, rather than a density whose width is read directly from the amplitude's exponent. With the unitary, negative-exponent convention for the Fourier transform, its momentum amplitude isThe Gaussian integral evaluates this Fourier transform. Consequently $|{\tilde{\psi }}_{\sigma }(p){|}^2$ is another normal distribution, with mean zero and standard deviationFor example, in dimensionless units with $\hslash =1$, $\sigma =1/2,1,2$ give $(\Delta x,\Delta p)=(1/2,1),(1,1/2),(2,1/4)$. The Gaussian wave packets all saturate the same bound: narrowing one density broadens the other. More precisely, choose a reference length $L$ and set $X=x/L$, $P=pL/\hslash$; then $[X,P]=iI$, and the plotted widths are $\sigma /L$ and $L\Delta p/\hslash$.

The Heisenberg uncertainty relation connects localization to the cost in kinetic energy. In a quantum harmonic oscillator with mass $m>0$ and angular frequency $\omega >0$, the variance contribution to the expectation value of the energy is bounded using the arithmetic-geometric mean inequality:The mean-dependent contributions are nonnegative. The Gaussian eigenstate for a quadratic potential with ${\sigma }^2=\hslash /(2m\omega )$ and zero means reaches this ground state energy. This is why the tradeoff is more than a limitation on preparation: it predicts an unavoidable positive minimum energy.

Heisenberg's original article appeared in March 1927 as "Über den anschaulichen Inhalt der quantentheoretischen Kinematik und Mechanik", Zeitschrift für Physik 43, pp. 172–198. Its abstract discusses simultaneous imprecision of canonically conjugate quantities and explanatory thought experiments. The precise standard deviation formulation here distinguishes preparation spreads from a slogan about measurement disturbance. ^[ref].

A coherent state is a normalized eigenstate of the annihilation operator, $a|\alpha ⟩=\alpha |\alpha ⟩$. Expanding in the Fock state eigenbasis gives a recurrence $c_{n+1}=\alpha c_n/\sqrt{n+1}$, and wavefunction normalization fixesThe displacement operator therefore shifts the oscillator ground state in phase space. Distinct coherent states are not orthogonal:

The occupation probabilities form a Poisson distribution,For the quantum harmonic oscillator, the expectation values of position and momentum areThe fluctuations equal those of the ground state,Thus the position wavefunction is a displaced Gaussian wave packet that saturates the Heisenberg uncertainty principle.

Time evolution under $H=\hslash \omega (a^{†}a+1/2)$ preserves this form:The center follows the classical oscillator trajectory while the packet width stays constant. Its mean energy is $\hslash \omega (N+1/2)$; it is generally a superposition of energy eigenstates, rather than a state of definite energy. For $N\gg 1$, the relative occupation fluctuation is $1/\sqrt{N}$, explaining the classical appearance of a highly excited coherent state.

Use the condensate wavefunction convention $\int |\psi {|}^2{d}^{3}r=N$, so $n(\mathbf{r})=|\psi (\mathbf{r}){|}^2$ is the number density and $\psi /\sqrt{N}$ is a normalized one-particle orbital. The contact interaction strength is $U_0=4\pi {\hslash }^{2}a/m$ for dilute atoms with s-wave scattering length $a$. The chemical potential $\mu =\partial E/\partial N$ fixes the stationary time dependence $e^{-i\mu t/\hslash }$ and enforces normalization when minimizing the energy. The Gross–Pitaevskii equation follows from an energy functional that contains the pair-counting factor $1/2$:Varying $E-\mu \int |\psi {|}^2{d}^{3}r$ gives the stationary Gross–Pitaevskii equation; the interaction contribution to that equation is $U_{0}n\psi$, without the factor $1/2$.

For the broadening discussed here assume repulsive interactions, $a>0$. The Thomas–Fermi approximation for a condensate neglects the gradient kinetic energy compared with the trapping and interaction energies. At fixed spatial scale, increasing $N{U}_0$ increases the mean-field potential. Expansion then reduces the kinetic energy further: for a cloud radius $R$, its scale per atom is ${\hslash }^2/(m{R}^2)$, whereas the trap energy scale is $m{\omega }^2{R}^2$. Their ratio decreases as $(a_0/R{)}^4$. In the weak-interaction limit the kinetic energy instead balances confinement, giving the Gaussian wave packet ground state of the quantum harmonic oscillator.

The spherical Thomas–Fermi condensate hasIntegrating this density gives $N=8\pi \mu R^3/(15U_0)$. ThusAt fixed $N$, stronger repulsion produces a wider, lower wavefunction: the Gaussian wave packet crosses over towards the square root of an inverted parabola, with an almost flat central region and a steeper outer drop. The density is the inverted parabola; the wavefunction is its square root. The exact wavefunction remains smooth and has a tail, since kinetic energy cannot be neglected near the Thomas–Fermi condensate edge rounding region. The healing length must be small compared with the cloud size in the bulk. Attractive interactions would not follow this large-$N$ expansion argument.

The original plot compares the noninteracting endpoint with two normalized repulsive spherical Thomas–Fermi condensate profiles. The latter are asymptotic shapes, rather than exact intermediate solutions of the Gross–Pitaevskii equation. Each has the same integrated atom number; $\chi =Na/a_0$ labels the illustrative interaction strengths.

Choose $A$ real and positive. The Gaussian integral givesThe three zero-point energies add for each atom. The ideal-gas chemical potential is a one-atom energy; the total energy includes all $N$ atoms.

For weak interactions, the first-order correction in perturbation theory is the energy functional of the contact interaction evaluated on the unperturbed Gaussian wave packet. The integral needed isConsequently the Gaussian interaction energy of a trapped condensate isHere $\hslash \omega ={\hslash }^2/(m{a}_0^2)$. These are the mean-field expressions consistent with the supplied equation and $\int |\psi {|}^2=N$. Exact first-order perturbation theory for precisely $N$ atoms in one orbital replaces $N^2$ by $N(N-1)$ through the number of distinct pairs; this distinction is negligible in the large-$N$ regime of the crossover estimate.

To judge the ideal-gas approximation, compare the interaction energy with the total ideal energy, or compare the interaction energy per atom with $\hslash \omega$:Thus the crossover from Gaussian to Thomas–Fermi condensate is governed by $\chi =Na/a_0$: perturbation theory is reliable for $\chi \ll 1$, while the regime of the Thomas–Fermi approximation for a condensate requires $\chi \gg 1$. The characteristic crossover number is $N_c\sim a_0/a$; there is no sharp critical atom number here. If one specifically defines crossover by $E_I=E_0$ using the weak-interaction expression, it gives $N_c=(3\sqrt{2\pi }/2)a_0/a$. That prefactor is a convention-dependent estimate extrapolated to the edge of perturbation theory, not a phase transition. Comparing $E_I$ itself to a single $\hslash \omega$ would incorrectly omit a factor of $N$.

For the numerical estimate use $m\simeq 87\times 1.66054\times 10^{-27}\mathrm{kg}$ and $\hslash \simeq 1.05457\times 10^{-34}\mathrm{J}\mathrm{s}$. ThenThe particular equality $E_I=E_0$ would instead give about $1.15\times 10^3$ atoms. These quantify different choices within a smooth crossover; a well-developed spherical Thomas–Fermi condensate profile requires going substantially beyond the characteristic scale.

For the proposed Gaussian wave packet, differentiation givesInserting these in the Time-dependent Schrödinger equation with Hamiltonian $(p^2-x^2)/2$, its right-hand side becomesEquating the constant and quadratic coefficients with $i\hslash {\partial }_t\Psi$ yieldsConversely these two equations make the substitution an identity, so they are sufficient for the Gaussian evolution in an inverted harmonic oscillator.

Write $B=b+ic$ with $b>0$. Normalization of the Gaussian wave packet gives $|A{|}^2\int e^{-2b{x}^2}dx=1$, hence $|A{|}^2=\sqrt{2b/\pi }$. The supplied Gaussian integral ratio then givesThe momentum operator obeys $p\Psi =2i\hslash Bx\Psi$, and direct second differentiation gives $p^2\Psi =(2{\hslash }^{2}B-4{\hslash }^2{B}^2{x}^2)\Psi$. Taking the expectation value and substituting the first result givessince $2b-B=\bar{B}$. These are the second moments of a complex Gaussian wave packet.

For the solution in part (b), the trigonometric identitygives, writing $D=\cosh (2t)+\cos (2\varphi )$,Thus normalizability requires $\sin (2\varphi )>0$, and the exact expectation values areSince $\cosh (2t)\sim e^{2t}/2$,Their difference is the constant $-\hslash \cot (2\varphi )$, so the mean energy $⟨H⟩=-\hslash \cot (2\varphi )/2$ stays constant despite exponential spreading in both observables.

For a normalized state and an observable $A$, define $⟨A⟩=⟨\psi ,A\psi ⟩$ and ${\Delta }_{\psi }A=\Vert (A-⟨A⟩)\psi \Vert$. Suppose first that the position and momentum variances are finite and the state has the regularity needed for the usual position-momentum quadratic forms. Put $a=⟨x⟩$, $b=⟨p⟩$, $u=(x-a)\psi$ and $v=(-i\hslash {\partial }_x-b)\psi$. The Cauchy-Schwarz inequality givesThe term involving $b$ has zero imaginary part. Integration by parts givesConsequently $\Im ⟨u,v⟩=\hslash /2$, andThis proves the Heisenberg uncertainty relation. For smooth rapidly decaying states the boundary terms vanish directly. The same proof extends to $x\psi \in L^2$ and weak derivative ${\psi }^{\prime }\in L^2$ using smooth cutoffs and taking their radius to infinity; the necessary mixed products are integrable by Cauchy-Schwarz. Normalization alone need not give finite variances or an operator-domain momentum expectation. A state with infinite spread is interpreted through the corresponding extended bound, rather than assigning a finite uncertainty to an undefined expression.

For the specified Gaussian wave packet, write $s=\hslash t$ and $a_t=1+is$. Its probability density isIt is a normalized centered normal distribution with variance $1+s^2$. Symmetry gives $⟨x⟩=0$, and henceAlso ${\partial }_x\psi =-x\psi /(2a_t)$, so $p\psi =i\hslash x\psi /(2a_t)$ and $⟨p⟩=0$ by oddness. Integration by parts, or directly taking the squared norm, givesThusThe packet saturates the bound at $t=0$ and has a larger position-momentum uncertainty product at every nonzero time. Its spatial spreading does not change the momentum distribution.

A free nonrelativistic Gaussian wave packet with initial position variance ${\sigma }_0^2$ has wave functionThe free dispersion relation $E=p^2/(2m)$ makes its different momentum components acquire different phases. Fourier evolution gives a position varianceThis spreading is ordinary unitary dynamics. A Gaussian already has nonzero tails initially, so its width increase alone is not an example of information suddenly appearing outside an initially compact support.

For genuinely compactly supported nonzero initial data, the nonrelativistic free kernel yieldsAt $t\ne 0$, the integral is an entire Fourier transform of a compactly supported function. It cannot vanish on an open exterior interval unless the whole transform, and hence the initial state, vanishes. Thus this theory has no strict finite propagation cone for nonzero compact initial wave functions. It is a nonrelativistic theory, so that result is not a contradiction in a fundamental relativistic field theory.

Positive-energy relativistic particle descriptions raise a subtler issue. The Hegerfeldt theorem concerns a Hamiltonian bounded below and a positive localization effect $A$. The function $p_A(t)=\Vert A^{1/2}{e}^{-iHt}\psi {\Vert }^2$ either vanishes identically or is nonzero for almost every time. Vanishing for an open interval makes the analytic continuation of the vector matrix elements vanish identically. A sharp exterior localization probability which remains zero for a nonzero time interval but becomes positive later is therefore incompatible with those assumptions.

The conclusion restricts simultaneous assumptions about positive energy and sharp particle localization. It does not establish an admissible procedure for controllable superluminal messaging. A particle-position projection need not belong to the algebra of a physically local relativistic measurement, and strictly localized one-particle preparation can fail the required assumptions. In field theory the appropriate locality test concerns local operations and observables, not merely pointwise values of a one-particle wave function. Localization, propagation of field disturbances and transmission of information must be distinguished.

Put $D=1+i\hslash t/m$ and write the Gaussian wave packet as $\psi ={\pi }^{-1/4}{D}^{-1/2}{e}^{-x^2/(2D)}$. Differentiation givesHence $\boxed{i\hslash {\psi }_t=-({\hslash }^2/2m){\psi }_{xx}}$, the free Time-dependent Schrödinger equation.

With $\tau =\hslash t/m$, its probability density isThe stated Gaussian integral shows its integral is one. Symmetry gives $⟨x⟩=0$; differentiating that Gaussian integral with respect to its coefficient gives $⟨x^2⟩=(1+{\tau }^2)/2$. Thus the position standard deviation isIt is not a stationary state: its probability density changes with time as the packet spreads. A stationary state would have only an overall time-dependent phase and a fixed position density.

The momentum operator in the position representation is $\hat{p}=-i\hslash d/dx$. Differentiating the given Gaussian wave packet givesThe probability density $|\Psi {|}^2$ is an even function, so the term proportional to the odd function $x$ has integral zero. Normalization supplies ${\int }_{-\infty }^{\infty }|\Psi {|}^{2}dx=1$, and hence

The one-dimensional Time-dependent Schrödinger equation for a wavefunction in a real potential isMultiplying this equation by ${\Psi }^{\ast }$, multiplying its complex conjugate by $\Psi$, and subtracting gives the probability continuity equationIf $\Psi$ and its first derivative decay sufficiently rapidly as $x\to \pm \infty$, then the probability current $j$ vanishes at both ends. Integrating the continuity equation and using the fundamental theorem of calculus yieldsThis conservation of quantum probability is required by the Born rule: once a normalizable wavefunction has total probability one, its time evolution must preserve that normalization.

For the stated Gaussian wave packet, put $\beta =\hslash t/m$. Sinceits squared modulus isThe Gaussian integral then giveswhich is independent of time, as required.

An initially localized escaping electron occupies a superposition of lead momentum states. In a uniform free one-dimensional lead,A packet centered at $k_0$ moves with group velocity $\hslash k_0/m^{\ast }$ and generally spreads because its components have different velocities. For an initially unchirped minimum-uncertainty Gaussian wave packet,This illustrates dispersive evolution; the actual emission contact determines $a(k)$ and whether only outward-propagating components are appreciable. It does not instantaneously become a single lead plane wave.

A single-electron transistor used as a sensitive electrometer exploits Coulomb blockade: a nearby charge shifts the offset $Q_{\mathrm{ext}}$ and therefore the conductance peaks. This requires stable calibrated capacitances and low charge noise, rather than merely a small conductor.

Put $s=1+it$. Logarithmic differentiation givesThereforeso the appropriate potential energy is $\boxed{V(x)=0}$. The wavefunction never vanishes, so no other time-independent potential gives the same equation for this precise phase choice.