How to prove that a linear operator acting on a null ket produces a null ket?

Question

Help me solve this question. Isn't it obvious that when a linear operator, say $A$, acts on a null ket, it will give a null ket? That's like a property of linear operator. How do I prove this?

Answer 1

Let the null ket be denoted $ \newcommand{\p}[2]{\frac{\partial #1}{\partial #2}} \newcommand{\f}[2]{\frac{ #1}{ #2}} \newcommand{\l}[0]{\left(} \newcommand{\r}[0]{\right)} \newcommand{\mean}[1]{\langle #1 \rangle}\newcommand{\e}[0]{\varepsilon} \newcommand{\ket}[1]{\left|#1\right>} \newcommand{\bra}[1]{\left<#1\right|} \newcommand{\braoket}[3]{\left<#1\right|#2\left|#3\right>} \ket{0}$ and let an arbitrary ket be $\ket{A}$ the property of the null ket is that: $$\ket{0}+\ket{A}=\ket{A}$$ This is the property we can use to find the action of the linear operator, $\hat B$, on the null ket. Since we must have: $$\hat B\ket{0}+\hat B\ket{A}=\hat B\ket{A}$$ Let $\ket{A'}=\hat B\ket{A}$ which gives: $$\hat B\ket{0}+\ket{A'}=\ket{A'}$$ I will leave it to you to decide why this implies $\hat B\ket{0}$ is a null ket.