In quantum mechanics, the behavior of particles is described by wave functions that satisfy the Schrödinger equation. The Schrödinger equation for a particle of mass m in a one-dimensional potential V(x) is given by:

Ĥψ(x) = Eψ(x)

where Ĥ is the Hamiltonian operator, ψ(x) is the wave function, E is the energy of the particle, and x is the position.



The given potential is V(x) = ∞ for x < 0.="" this="" means="" that="" the="" potential="" is="" infinite="" for="" all="" positions="" with="" negative="" x.="" in="" other="" words,="" the="" particle="" is="" confined="" to="" the="" region="" x="" ≥="" />



In quantum mechanics, wave functions must be normalized, meaning that their integral over all space must be equal to 1. Additionally, wave functions must be continuous and have continuous derivatives.



The first excited state corresponds to the lowest energy state above the ground state. To find the energy of the first excited state, we need to solve the Schrödinger equation for the given potential.

Since the potential is infinite for x < 0,="" the="" wave="" function="" must="" be="" zero="" in="" this="" region.="" therefore,="" the="" wave="" function="" for="" the="" first="" excited="" state="" can="" be="" written="" />

ψ(x) = 0 for x < />

For x ≥ 0, the Schrödinger equation becomes:

-ħ²/(2m) d²ψ(x)/dx² + V(x)ψ(x) = Eψ(x)

Since V(x) = ∞ for x < 0,="" the="" equation="" simplifies="" />

-ħ²/(2m) d²ψ(x)/dx² = Eψ(x)

This is a second-order linear ordinary differential equation. The general solution to this equation is a linear combination of sine and cosine functions:

ψ(x) = A sin(kx) + B cos(kx)

where A and B are constants, and k is the wave number given by:

k = √(2mE)/ħ

Since the wave function must be zero for x < 0,="" the="" cosine="" term="" is="" not="" present="" in="" the="" solution.="" therefore,="" the="" wave="" function="" for="" the="" first="" excited="" state="" />

ψ(x) = A sin(kx)

To determine the energy E, we can substitute the wave function into the Schrödinger equation and solve for k. The boundary condition ψ(0) = 0 gives:

0 = A sin(0)

This implies that A = 0, which means the wave function is identically zero for all x. Therefore, the first excited state does not exist in this potential. The energy of the first excited state is therefore undefined.



In a one-dimensional potential where the potential is infinite for x < 0,="" the="" first="" excited="" state="" does="" not="" exist.="" the="" energy="" of="" the="" first="" excited="" state="" is="" undefined.="" this="" result="" is="" due="" to="" the="" infinite="" potential="" barrier="" at="" x="0," which="" prevents="" the="" 0,="" the="" first="" excited="" state="" does="" not="" exist.="" the="" energy="" of="" the="" first="" excited="" state="" is="" undefined.="" this="" result="" is="" due="" to="" the="" infinite="" potential="" barrier="" at="" x="0," which="" prevents="" />