# Free particle in Quantum mechanics

Recently I am studying R.Shankar Principles of Quantum mechanics. I started with the fifth chapter(I did cover the previous part so that I had good context). I got stuck at one of the parts in the derivation for a free particle. The book started of by writing the Schrodinger wave equation:$$\frac{P^2}{2m}|\psi\rangle=i\hbar \frac{d}{dt}|\psi\rangle$$ and then,$$|\psi\rangle =|E\rangle e^{\frac{-iEt}{\hbar}}$$ Where E is the stationary state solution. To find $$|E\rangle$$ we solve the Time-independent Schrodinger wave equation:$$\frac{P^2}{2m}|E\rangle =E|E\rangle$$ and now solving it in the Momentum basis with reduce the equation to:$$\left[\frac{p^2}{2m}-E\right]|\psi\rangle=0$$ and thus $$p=\pm\sqrt{2mE}$$ and then the book say's that the ket $$|E\rangle$$ is:$$|E\rangle=\beta|p=+\sqrt{2mE}\rangle +\gamma|p=-\sqrt{2mE}\rangle$$ I didn't understand how they were able to make such a statement? When I dig a bit more in the maths I realised that $$p$$ is the eigenvalue:$$P|p\rangle=p|p\rangle$$ I read carefully and found a statement: "Any eigenstate of $$p$$ is also an eigenstate of $$p^2$$".I wanted to prove this.
I started with:$$P^2|\alpha\rangle=\alpha|\alpha\rangle$$$$P|\beta\rangle=\beta|\beta\rangle$$In the end I reached:$$\langle \beta|\alpha\rangle(\beta^2-\alpha)=0$$ And I got stuck. I got even more confused about why is $$|E\rangle$$ is what is it? It will be really helpful if you can help. I am a newcomer to Quantum mechanics and struggling a bit with the imagination problem of functions as vector.

• My main question is what is happening in the Momentum basis?

It will be really helpful If you can help.

• This question-and-answers might be helpful. Commented Dec 20, 2023 at 17:19
• Where is the problem, exactly? You realized that $P|p\rangle=p|p\rangle$. Then $P^2=P\circ P$, i.e. it means "apply $P$ twice", giving $P^2|p\rangle =P\circ P |p\rangle=P (p|p\rangle)=p P|p\rangle = p^2|p\rangle$. Commented Dec 20, 2023 at 17:57
• @TobiasFünke I understood it. My question is that how they constructed the $|E\rangle$ from p? there was no maths. Commented Dec 20, 2023 at 18:14
• A free particle only has kinetic energy. $$E=\frac12mv^2=\frac{(mv)^2}{2m}=\frac{p^2}{2m}$$ Commented Dec 21, 2023 at 0:56

That's a good question. Actually, you can get solutions of Schrödinger's equation to negative energy (which therefore don't belong to any momentum). However, not even linear combinations of those solutions can be normalized. That's why those solutions can be considered unphysical. You can find that kind of argument in many textbooks, for example Griffiths (if I remember correctly). Another argument is that in general common eigenstates of commuting observables form a basis, so it's okay to just consider the eigenstates of both the hamiltonian and the momentum operator. That trick is often used in quantum mechanics, for example also in the case of angular momentum or the hydrogen atom. If you want to understand this mathematically, you'll have to get into spectral theory. Obviously, $$e^{i p \cdot x/\hbar}$$ is not an eigenstate of the momentum operator in the normal sense, it's not even part of Hilbert space. Therefore, you have to introduce generalized eigenstates and if you do that properly, you get that negative energy states are not such generalized eigenstates. The argument that commuting observables have bases of common eigenstates is also true in "mathematical" functional analysis (I think).