Why is probabilty conserved under time evolution of a system in quantum mechanics?

I've studied quantum mechanics to a certain degree, but one question that I've never been able to get a fully satisfactory answer to is why probability is conserved (by this I mean that it has either just been stated to me that the norm is preserved under time evolution, or through taking the time derivative of the bra-ket and showing it to be zero through use of the schroedinger equation, which is fine mathematically, but I would really like an intuitive physical reason also). Or, in other words, why is the norm of a state vector conserved, i.e. $$\big\vert\langle\psi\vert\psi\rangle\big\vert = 1 \qquad\forall\;t$$ Is the intuitive physical reasoning that whatever observable we are measuring, the total probability that it has a value (within it's allowed range) should always be unity (e.g. it must always have some value of momentum, position etc.)?!

• Since you've stated that you've found no fully satisfactory answer, it would be appropriate, I think, for you to precisely identify in your question what isn't satisfactory so those here that wish to answer your question will not have to guess at what doesn't satisfy you. – Alfred Centauri Apr 22 '15 at 12:20
• If we constrain a particle to some region of space, we should always find it in that region. Thus, we should expect the probability over all space be 1 for all time; simple as that. – Kyle Kanos Apr 22 '15 at 13:59
• Related: physics.stackexchange.com/q/169936/2451 and links therein. – Qmechanic Apr 22 '15 at 15:57

Mathematical reason is that the time evolution operator is unitary, which means that $U^\dagger = U^{-1}$. Therefore $\langle \psi(t) | \psi(t) \rangle = \langle \psi(0)| U^\dagger U | \psi(0) \rangle = \langle \psi(0) | \psi(0) \rangle$. We can see that it is unitary by considering the Schrodinger equation: $$\newcommand{\ket}{| #1 \rangle} \frac{\partial}{\partial{t}} \ket{\psi(t)} = \frac{1}{i \hbar} \hat{H} \ket{\psi(t)}$$ Let's assume the Hamiltonian is time-independent. (We don't need to but it makes things easier.) And let's assume there is a linear operator $U(t)$ that time evolves kets. Then we can write: $$\frac{\partial}{\partial{t}} U(t) \ket{\psi(0)} = \frac{1}{i \hbar} \hat{H} U(t)\ket{\psi(0)}$$ But this needs to hold for arbitrary $\ket{\psi(0)}$, so the operators must be equivalent: $$\frac{\partial}{\partial{t}} U(t) = \frac{1}{i \hbar} \hat{H} U(t)$$ This solution to this differential equation is: $$U(t) = e^{-i \frac{\hat{H}}{\hbar} t}$$ The initial condition here is the reasonable one, that $U(0) = 1$. Then we can ask ourselves what the Hermitian conjugate of this is. Since $\hat{H}$ is Hermitian, it follows: $$U^\dagger(t) = e^{i \frac{\hat{H}}{\hbar} t}$$ And from that we can see that $$U(t) U^\dagger(t) = e^{i \frac{\hat{H}}{\hbar} (t - t)} = 1\\ \implies U^\dagger(t) = U^{-1}(t)$$ Notice that the time evolution operator's inverse is just the evolution operator for going to $-t$ rather than $+t$, so the unitarity is an expression of time-reversal symmetry.
Another reason might be to consider what a probability is. It doesn't really make sense to say that the probability won't add up to 1. Even if you say that the momentum could be zero, that's still some probability in $| p = 0 \rangle$. How would the particle simply not have a measured value of momentum? An expectation value maps the wavefunction to a number. Something very odd is going on if you don't get that number, and the nature of what it is to be a probability means that it should add to precisely 1.
At the risk of complicating things, in some applications people use non-unitary time evolution. For instance, in scattering calculations we might want to include an imaginary potential, which can absorb probability. This could correspond to the fact that my particle might not come back from the scattering event--maybe it's captured or something. However, this is mostly a computational trick--if I included the bound state that the particle could go into, I should still have unitary time evolution, but maybe some of my probability ends up in the bound state instead of the continuum of $| k \rangle$ states. The lack of probability tells us that something has escaped our analysis, not that the true expectation value is undefined. If you do this, you do it (1) on purpose and (2) with a clear prescription of how to handle the probabilities for the wavefunction you have left over. For instance, to actually find an expectation value of the scattered wavefunction which may have lost some amplitude, you need to--you guessed it--rescale so probability is again 1.
• Yes. Imagine the particle traveling through space--it moves, so the probability as a function of position $P(x)$ is changing, but it's always somewhere, so the probability should always sum to 1. – zeldredge Apr 22 '15 at 14:13
• Consider a time-independent Hamiltonian. Then we can show that the evolution operator is $exp (-i \hat{H} t )$. If $\hat{H}$ is Hermitian, then this will be unitary. (For time-dependent Hamiltonians, this also holds, but you have to use the infinitesimal operator. I can show you that if you feel skeptical still.) Reference: Cohen-Tahnnoudji, "Quantum Mechanics", Complement F-III – zeldredge Apr 22 '15 at 14:24