Understanding dot product in quantum mechanics Let's say we have a two-state-system with state $\vert 1\rangle$ and state $\vert 2\rangle$. From my understanding one can assume the base vectors of this system to be $\vert1\rangle \mapsto (1,0)^\top$ and $\vert 2\rangle \mapsto (0,1)^\top$.
Now assume we have a particle in state $\vert \phi\rangle = (\sqrt{\frac{1}{3}}, \sqrt{\frac{2}{3}})^\top$ and a different state $\vert \psi\rangle = (\sqrt{\frac{1}{2}}, \sqrt{\frac{1}{2}})^\top$. $\langle\psi|\phi\rangle$ should now define the amplitude to go from $\vert \phi\rangle$ to $\vert \psi\rangle$. This can be rewritten as $\langle\psi|1\rangle\langle 1|\phi\rangle + \langle\psi| 2\rangle\langle2|\phi\rangle$.
So putting in the numbers we get
$$\langle\psi|\phi\rangle = \sqrt{\frac{1}{2}}\cdot\sqrt{\frac{1}{3}} + \sqrt{\frac{1}{2}}\cdot\sqrt{\frac{2}{3}} \approx 0.985.$$
Is this thinking correct? Can a particle without any influence even change its state? (I suppose yes, as it would be otherwise in a stable state, which are only a small subset of the vectors in the vector space).
 A: The inner (dot) product is completely correct and all, though to get a probability you'll want to square its modulus, so $|\langle \psi | \phi \rangle |^2$.
The interpretation of the number $0.985^2$ depends on the context though. The relevant postulate in QM is

When a particle is in state $|\phi \rangle$ and a measurement is made on some observable with corresponding operator $\hat{p}$ which has eigenvectors $|p\rangle$ and corresponding eigenvalues $p$, then the probability of getting a result equal to $p$ in that measurement is $|\langle p | \phi \rangle | ^2$.

So, in order to interpret your inner product squared as a probability, it first has to be the case that you are in a physical scenario in which the state $|\psi \rangle$ is an eigenvector of the operator corresponding to a quantity you measured. Outside of a specified measurement scenario, that squared inner product does not carry any special meaning and is just a random number.
For that reason you can't interpret it in general like "the state has a probability of randomly transitioning to $|\psi \rangle$ with probability $0.985^2$" without further context. This doesn't just happen willy-nilly, these transition/jumps occur as a result of someone measuring the particle.
Edit: Swapped the definition of $\psi$ and $\phi$ to be consistent with the question.
A: When you are talking about going from a state $|\Psi\rangle$ to a state $|\Phi\rangle$ you are implicitly talking about a perturbative process. Going in this direction, "can a particle (without influence) change state?" is a natrual question to ask.
As I said, when talking about transition between states, you are in the perturbative regime. That means if your system is described by a Hamiltonian $H_0$ of which you know eigenstates and energies, lets call them $|n\rangle$ with energies $E_n$ and your system is initally in some state $|m\rangle$, then your system will always be in this state. So there will never be a transition between eigenstates.
But if your system is more accurately described by the Hamiltonian $H = H_0 + V$, where $V$ is a small perturbing potential and can also depend on time (small here is a condition for perturbation theory. Just look it up in your favourite QM book). Then you can, under certain conditions, describe your system approximately with your states $|n\rangle$ to zeroth order. But if you now have an initial state $|m\rangle$, your system will not stay in this state, but will transition with a certain probability per time into different states approximately described by your unperturbed eigenstates $|n\rangle$ (note that now the system does not stay in $|m\rangle$, since it is not an exact eigenstate of $H$, which governs the time evolution of your system).
One can calculate this transition rate with Fermi's Golden Rule, namely the probability per unit time to go from state $|i\rangle$ to state $|f\rangle$ is
$$\Gamma_{i\rightarrow f} = \frac{2\pi}{\hbar}|\langle i | V|f\rangle|^2 \rho(E_f),$$
where $\rho(E)$ is the density of states.
This result comes with all sort of assumptions, so if you want more information, again just look into a QM book.
EDIT:
In your example when you talk about the matrix element $\langle \Psi |\Phi\rangle$, you basically have $V=id$. This is a special kind of perturbation, because it is diagonal in the eigenbasis of $H_0$. In this case the perturbation will never cause a transition, since $\langle \Psi |\Phi\rangle$=0 for basis states $|\Psi\rangle$ and $|\Phi\rangle$ (when talking about eigenstates, one usually means elements of the eigenbasis of the Hilbert space with resepect to $H_0$).
