\(\left\langle {\Psi}\middle|{\Psi}\right\rangle =1\) Identify and discuss the dimensions of \(\left|{\Psi}\right\rangle \).
Taking the Hermitian adjoint of a ket does not change the dimensions/units of the ket, so since the bracket is dimensionless, the ket \(\left|{\psi}\right\rangle \) is also dimensionless.
For a spin-\(\frac{1}{2}\) system, \(\left\langle {\Psi}\middle|{+}\right\rangle \left\langle {+}\middle|{\Psi}\right\rangle + \left\langle {\Psi}\middle|{-}\right\rangle \left\langle {-}\middle|{\Psi}\right\rangle =1\). Identify and discuss the dimensions of \(\left|{+}\right\rangle \) and \(\left|{-}\right\rangle \).
Similar to the previous part, each term of the left hand side must be dimensionless. Therefore, the spin states \(\left|{+}\right\rangle \), and \(\left|{-}\right\rangle \) are both dimensionless.
In the position basis \(\int \left\langle {\Psi}\middle|{x}\right\rangle \left\langle {x}\middle|{\Psi}\right\rangle dx = 1\). Identify and discuss the dimesions of \(\left|{x}\right\rangle \).
Here, the integral is dimensionless, but the \(dx\) introduces dimensions of length. We already know the \(\left|{\Psi}\right\rangle \) is dimensionless, so \(\left|{x}\right\rangle \) must have dimensions of inverse square root length, \(1/\sqrt{L}\).
\[\int \left\langle {\Psi}\right|\underbrace{\left|{x}\right\rangle }_{(\frac{1}{\sqrt{L}})}\underbrace{\left\langle {x}\right|}_{(\frac{1}{\sqrt{L}})}\left|{\Psi}\right\rangle \underbrace{dx}_{L} = 1\]