# What is the probability of measurement in QM, dependent on time?

Consider a QM system with an observable $$A$$ and orthonormal eigenbasis $$\{|n\rangle,n=0,1,2,\ldots\}$$. Then we know that if the system is in some state $$|\psi\rangle$$ and we measure $$A$$, the probability of finding an eigenstate $$|n\rangle$$ is $$|\langle n|\psi\rangle|^2$$ and that the probability of finding a non-eigenstate $$|\phi\rangle$$ is zero.

How does this translate to the time-dependent setting?

In particular, what is the probability of finding a state $$|\phi,t\rangle$$ at time $$t$$ if the system evolves according to some other state $$|\psi,t\rangle$$? Does $$|\phi,0\rangle$$ have to be an eigenstate of $$A$$ in order for this probability to be positive? Does the time evolution of $$|\phi,t\rangle$$ have to be given by the Schrödinger equation? I'm really confused by this and would very much appreciate help.

• The probability of measuring a non-eigenstate $|\phi\rangle$ is $|\langle \phi|\psi\rangle|^2$, not zero. Commented Jul 21, 2021 at 10:37
• Commented Jul 21, 2021 at 11:27
• @Raskolnikov Thanks but I thought the system actually HAS to be in an eigenstate after measurement, or am I confusing things? Commented Jul 21, 2021 at 11:43
• @test123 Yes, the system has to be in an eigenstate after the wave function collapsed due to the measurement and yes, you are confusing things. If you have another experiment with which you can measure a non-eigenstate of the operator considered before, then the probability of that state will not be 0 unless the system is in a state orthogonal to the non-eigenstate. For example, a system can have non-zero probability for spin in x- and y-direction, but when measuring spin in x-direction, you can't get the result of spin in y-direction.
– sim0
Commented Jul 21, 2021 at 12:07

In quantum mechanics, each state of a system is always represented as a vector (a ket) in the Hilbert space of all possible states. This space has, per definition, a scalar product and thus a geometry associated with it and consequentially there is the notion of orthogonality. This allows for the probability of measuring any state $$|\phi\rangle$$ when the system is in state $$|\psi\rangle$$ to be defined as the square of the orthogonal projection $$|\langle \phi | \psi \rangle|^2$$, which is the scalar product. This definition is universal and does always apply. In particular, none of the states has to be an eigenstate of any operator.
Now to answer your question explicitly, by what I just explained, the probability of finding a state $$|\phi,t\rangle$$ at time $$t$$ if the system is in the state $$|\psi,t\rangle$$ is $$p = |\langle \phi,t | \psi,t \rangle |^2~,$$ and there are no special conditions which $$|\phi,0\rangle$$ must fulfill. Furthermore, it will be $$p > 0$$, if $$|\psi,t\rangle$$ and $$|\phi,t\rangle$$ are not orthogonal, by definition of orthogonality.