# What happens to the normalization condition if the wave function is non stationary?

The quantum mechanics courses I have taken as of now pretty much only deal with stationary states of the form:

$$\psi(x,t) = \psi(x)\exp{\left(-\frac{iE}{\hbar}t\right)}$$

Now, what about situations where time dependence is not limited to a phase factor? What if, say, the time dependence makes it so that $$\psi$$ tends to decrease? Wouldn't that mean that $$\int_{-\infty}^{\infty}|\psi|^2dx<1$$ as $$t>0$$?

• Hint: The time-evolution is unitary. Jan 30, 2023 at 19:49
• Hint: you surely have evolved the flipflop oscillator state $N(e^{-i\omega t/2}|0\rangle + e^{-i3\omega t/2}|1\rangle)$. How did you normalize it? Jan 30, 2023 at 20:04

The people that have developed quantum mechanics have indeed thought about this. I can show you that if you start with a normalized state then this will be normalized for all of time.

Consider the eigenstates $$\psi_n(x)$$ of the Hamiltonian. The states which satisfy $$\hat H\psi_n(x)=E_n\psi_n(x)$$ Under certain conditions$$^*$$ these states form an orthonormal basis. That is, they statisfy \begin{align} \langle\psi_m|\psi_n\rangle&=\int\mathrm dx\,\psi_m^*(x)\psi_n(x)\\&=\delta_{mn}\\&=\cases{1&m=n\\0&m\neq n} \end{align} If you view these states as vectors and view $$\langle\psi_m|\psi_n\rangle$$ as a generalized dot product then each state is orthogonal to each other state.

We have to use one more fact to show the probability is conserved. If these states form a complete basis we can express any function as a sum of these eigenstates: $$\psi(x)=\sum_nc_n\psi_n(x)$$

To normalize $$\psi$$ we have to normalize the sum of the coefficients. \begin{align} \langle\psi|\psi\rangle&=\int\mathrm dx\,\left(\sum_mc_m^*\psi_m^*(x)\right)\left(\sum_nc_n\psi_n(x)\right)\\ &=\sum_m\sum_nc_m^*c_n\int\mathrm dx\,\psi_n(x)\psi_m^*(x)\\ &=\sum_m\sum_nc_m^*c_n\delta_{mn}\\ &=\sum_n |c_n|^2\overset{\text{should be}}{=}1 \end{align}

If we apply time evolution to one of these eigenstates then the state just gets multiplied with a phase factor. Since the Schrödinger equation is linear, if we apply time evolution to a sum of eigenstates each eigenstate gets its own phase factor. $$\psi(x,t)=\sum_nc_n\psi_n(x,0)\exp\left(-iE_nt/\hbar\right)$$

Using orthonormality we can finally show that if we properly normalized our sum at $$t=0$$ the state will then be normalized forever.

\begin{align} \langle\psi(t)|\psi(t)\rangle&=\int\mathrm dx\,\left(\sum_mc_m^*\psi_m^*(x,0)\exp\left(iE_mt/\hbar\right)\right)\left(\sum_nc_n\psi_n(x,0)\exp\left(-iE_nt/\hbar\right)\right)\\ &=\sum_m\sum_nc_m^*c_ne^{-i(E_n-E_m)t/\hbar}\int\mathrm dx\,\psi_n(x)\psi_m^*(x)\\ &=\sum_m\sum_nc_m^*c_ne^{-i(E_n-E_m)t/\hbar}\delta_{mn}\\ &=\sum_n |c_n|^2=1 \end{align} You should note that for $$m=n$$ the phase factor becomes 1.

$$*$$ The condition is that the Hamiltonian is a Hermitian operator. In case you have not encountered this term you soon will. One property of Hermitian operators is that they ensure the eigenvectors are all real numbers, which is nice because the eigenvalues represent the energy.