# What is the wave function for states n=1,2,3 for a particle with m mass which is in infinite square well potential(x=-L/2, x=L/2)?

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You need to use the following equation of the wave function in an infinite square potential, with walls `x = -L/2, x = L/2` , such that:

Replacing -`L/2` for x yields:

`psi(-L/2) = sqrt2/L*sin ((n*pi*(-L/2 + L/2))/L)`

`psi(-L/2) = sqrt2/L*sin 0 => psi(-L/2) = sqrt2/L*0 => psi(-L/2) = 0`

Replacing `L/2` for x yields:

`psi(L/2) = sqrt2/L*sin ((n*pi*(L/2 + L/2))/L)`

`psi(L/2) = sqrt2/L*sin ((n*pi*L)/L)`

`psi(L/2) = sqrt2/L*sink*pi => psi(L/2) = sqrt2/L*0 = 0`

Since `psi(-L/2) = psi(L/2) = 0,` you may evaluate the state function at `n = 1, n = 2, n = 3` , such that:

`n = 1 => psi(x) = sqrt2/L*sin ((pi*(x + L/2))/L)`

`n = 1 => psi(x) = sqrt2/L*sin ((pi*x)/L + pi/2) = (sqrt2/L)(sin((pi*x)/L)cos(pi/2) + cos((pi*x)/L)sin(pi/2) ) = sqrt2/L*cos((pi*x)/L)`

`n = 2 => psi(x) = sqrt2/L*sin ((2pi*x)/L + pi) = sqrt2/L*(sin((2pi*x)/L)cos pi + cos((2pi*x)/L)sin pi) = -sqrt2/L*sin((2pi*x)/L)`

`n = 3 => psi(x) = sqrt2/L*cos((3pi*x)/L)`

**Hence, evaluating the wave function for states `n=1,n=2,n=3` , yields **`psi(x) = sqrt2/L*cos((pi*x)/L), psi(x) = -sqrt2/L*sin((2pi*x)/L), psi(x) = sqrt2/L*cos((3pi*x)/L).`