Applying angular momentum operator [closed]

How are the algebraic steps to applied the angular momentum operator defined as: $$\hat{L}=-i\hbar[r\times\nabla]$$ to $$\Psi=a~ \psi_{431}$$ where the $\psi_{nlm}$ are the eigenfunctions of the time independent Schrödinger equation for the hydrogen atom.

Note 1 : For the $\:\psi_{nlm}\:$ eigenfunction \begin{align} H\,| \psi_{nlm}\rangle & = E_{n} | \psi_{nlm}\rangle \tag{01}\\ L^{2}\,| \psi_{nlm}\rangle & = \ell\left(\ell+1\right)\hbar^{2} | \psi_{nlm}\rangle \tag{02}\\ L_{z}\,| \psi_{nlm}\rangle & = m\hbar | \psi_{nlm}\rangle \tag{03} \end{align} Note 2 : For information (not necessary) \begin{align} E_{n} & = \dfrac{E_{0}}{n^2}\tag{04a}\\ E_{0} & = \textrm{ground state energy}=-13.6~\textrm{eV} \tag{04b} \end{align}

closed as off-topic by Bill N, Gert, user36790, user10851, ACuriousMind♦Jun 30 '16 at 12:30

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• Do you not know how to calculate a gradient and cross product? – Ryan Unger Jun 29 '16 at 17:25
• what is $a$? The hydrogen wavefunctions are eigenfunctions of $\hat{L}$ – By Symmetry Jun 29 '16 at 17:29
• a it is a real constant. – ICAM Jun 29 '16 at 17:32
• @BySymmetry $\psi_{431}$ is an eigenstate of $L_z$, and ICAM is asking about the effect of the vector angular momentum operator. That said, there are clear rules for the interaction of each angular momentum component operator with eigenstates of $L_z$, and these are explained in detail in any suitable QM textbook. – Emilio Pisanty Jun 29 '16 at 17:45
• Note 1 : For the $\:\psi_{nlm}\:$ eigenstate \begin{align} H\,| \psi_{nlm}\rangle & = E_{n} | \psi_{nlm}\rangle \tag{01}\\ L^{2}\,| \psi_{nlm}\rangle & = \ell\left(\ell+1\right)\hbar^{2} | \psi_{nlm}\rangle \tag{02}\\ L_{z}\,| \psi_{nlm}\rangle & = m\hbar | \psi_{nlm}\rangle \tag{03} \end{align} Note 2 : For information (not necessary) \begin{align} E_{n} & =\dfrac{E_{0}}{n^2}\tag{04a}\\ E_{0} & = \textrm{ground state energy}=-13.6 eV \tag{04b} \end{align} – Frobenius Jun 29 '16 at 19:15

Let $\:\mathbf{L}\:$ the vector angular momentum operator $$\mathbf{L} =\left( L_{x},L_{y},L_{z} \right) \tag{01}$$ The eigenfunction $| \psi_{n\ell m}\rangle\:$ is a common eigenstate of $\:H,L^{2},L_{z}$ and more precisely for our case is an eigenstate of $\:L_{z}\:$ with eigenvalue $\:m\:$, so $$L_{z}| \psi_{n\ell m}\rangle =m\hbar|\psi_{n\ell m}\rangle \quad \Longrightarrow \quad L_{z}| \psi_{431}\rangle =\hbar|\psi_{431}\rangle \tag{02}$$ Now, taking advantage of the definitions and properties of the raising and lowering operators we have
$$L_{\pm}| \psi_{n\ell m}\rangle =\left(L_{x} \pm i L_{y}\right)| \psi_{n\ell m}\rangle=\hbar \sqrt{\ell\left( \ell+1\right)-m\left( m \pm1\right)}| \psi_{n\ell \left( m\pm1\right)}\rangle \tag{03}$$ giving for our $| \psi_{431}\rangle\:$ $$L_{\pm}| \psi_{431}\rangle =\left(L_{x} \pm i L_{y}\right)| \psi_{431}\rangle=\hbar \sqrt{12-1\left( 1 \pm 1\right)}\,| \psi_{43\left( 1\pm1\right)}\rangle \tag{04}$$ so \begin{align} L_{+}| \psi_{431}\rangle & =\left(L_{x} + i L_{y}\right)| \psi_{431}\rangle=\hbar \sqrt{10}\,| \psi_{432}\rangle \tag{05+}\\ L_{-}| \psi_{431}\rangle & =\left(L_{x}- i L_{y}\right)| \psi_{431}\rangle=\hbar \sqrt{12}\,| \psi_{430}\rangle \tag{05 - } \end{align} From (5$\pm$) \begin{align} L_{x}| \psi_{431}\rangle & =\:\:\:\dfrac{1}{2}\:\:\,\left(L_{+} + L_{-}\right)| \psi_{431}\rangle=\:\:\:\:\hbar \,\biggl(\sqrt{\dfrac{5}{2}}|\psi_{432}\rangle+\sqrt{3}|\psi_{430}\rangle\biggr) \tag{06x}\\ L_{y}|\psi_{431}\rangle & = -\dfrac{1}{2}i\left(L_{+} - L_{-}\right)| \psi_{431}\rangle=-i\hbar \,\biggl(\sqrt{\dfrac{5}{2}}|\psi_{432}\rangle-\sqrt{3}|\psi_{430}\rangle\biggr) \tag{06y} \end{align} and finally $$\boxed {\bbox[#FFFF88,8px]{\mathbf{L}|\psi_{431}\rangle =\hbar\Bigg[ \biggl(\sqrt{\dfrac{5}{2}}|\psi_{432}\rangle+\sqrt{3}|\psi_{430}\rangle\biggr),-i\,\biggl(\sqrt{\dfrac{5}{2}}|\psi_{432}\rangle-\sqrt{3}|\psi_{430}\rangle\biggr),|\psi_{431}\rangle\Biggr]}} \tag{07}$$