Linearly indep. annihilation operator

[yakattack]

Homework Statement

If $$\varphi$$ and $$\widetilde{ \varphi }$$ are linearly independent and
$$\hat{N}$$$$\varphi$$=n$$\varphi$$ and
$$\hat{N}$$$$\widetilde{\varphi}$$=n$$\widetilde{\varphi}$$, with $$n\geq 1$$
$$\ \text{Prove that } \hat{a}\varphi_{n} \text{ and } \hat{a}\widetilde{\varphi}_{n} \text{ are also linearly independent.}$$

Homework Equations

$$\ \text{I have to use the fact that if } \hat{N}\varphi=\nu\varphi \text{ then } \nu\geq0 \text{ and } \nu=0 \text { iff } \hat{a}\varphi=0 \hat{N}=\hat{a^{*}}\hat{a} \hat{a} \text { is the annihilation operator.}$$

The Attempt at a Solution

If i suppose they aren't linearly independent then i can show that this means that
$$\hat{a}$$$$\varphi$$$$_{n}$$ and the same with tildas are linearly dependent. But does this show that the converse is true?

 

[Jhero]

Thank you for your post. To prove that \hat{a}\varphi_{n} and \hat{a}\widetilde{\varphi}_{n} are linearly independent, we can use the definition of linear independence. This means that we need to show that the only solution to the equation a\hat{a}\varphi_{n}+b\hat{a}\widetilde{\varphi}_{n}=0 is a=b=0.

First, let's consider the case where a=0 and b=0. In this case, we have 0\hat{a}\varphi_{n}+0\hat{a}\widetilde{\varphi}_{n}=0, which is true. Therefore, we have shown that a=b=0 is a solution to the equation.

Now, let's consider the case where a\neq 0 and b\neq 0. From the given information, we know that \hat{N}\varphi=n\varphi and \hat{N}\widetilde{\varphi}=n\widetilde{\varphi}. This means that \hat{N}\hat{a}\varphi=\hat{a}\hat{N}\varphi=\hat{a}(n\varphi)=n(\hat{a}\varphi) and similarly for \hat{a}\widetilde{\varphi}. This shows that \hat{a}\varphi_{n} and \hat{a}\widetilde{\varphi}_{n} are eigenstates of \hat{N} with eigenvalues n.

Now, suppose that a\hat{a}\varphi_{n}+b\hat{a}\widetilde{\varphi}_{n}=0. This means that \hat{a}(a\varphi_{n}+b\widetilde{\varphi}_{n})=0. Since \hat{a}\varphi_{n} and \hat{a}\widetilde{\varphi}_{n} are eigenstates of \hat{N}, this implies that a\varphi_{n}+b\widetilde{\varphi}_{n} is also an eigenstate of \hat{N} with eigenvalue n. However, since \varphi and \widetilde{\varphi} are linearly independent, this implies that