# Initial condition of the wave function while solving the Schrodinger equation

To calculate the evolution of a wave function under Schrodinger equation, we need to set an initial condition. In this context, can we associate a unique wave function to a given physical (observable) intial condition? The modulus-squared value of the wave function corresponds to the probability amplitude. So, it seems to me that we may associate random phases to the initial wave function at every point in space. These phases should not affect the values of physical observables at a future time (resulting from the time-evolved wave function). Am I correct? Can we make similar arguments about the spinor field evolving under Dirac equation which governs the evolution of four complex numbers? i.e. Can we associate four random phases to the intial spinor field at every point in space?

• So you are asking if, for example, $\Psi(x,0)=\psi(x)\cdot e^{i\phi(x)}$ is indistinguishable for any function $\phi(x)$? Mar 11, 2021 at 18:46
• Yes, that is exactly what I am asking. Mar 11, 2021 at 19:07

Consider the following two states describing a free particle in 1 dimension $$\begin{eqnarray} \psi_1(x,t) &=& \frac{1}{\sqrt{2\pi}} \exp\left[i \left(k x - \frac{\hbar k^2}{2m}t\right)\right] \\ \psi_2(x,t) &=& \frac{1}{\sqrt{2\pi}} \exp\left[i \left((k+\Delta k) x - \frac{\hbar (k+\Delta k)^2}{2 m}t\right)\right] \end{eqnarray}$$ Both $$\psi_1$$ and $$\psi_2$$ satisfy the Schrodinger equation $$$$H \psi_{1,2} = \frac{p^2}{2m}\psi_{1,2} = E_{1,2} \psi_{1,2} \implies -\frac{\hbar^2}{2m}\frac{d^2}{dx^2} \psi_{1,2} = E_{1,2} \psi_{1,2}$$$$ where for $$\psi_1$$, the energy $$E_1$$ is $$E_1 = \hbar \omega_1 = \frac{\hbar^2 k^2}{2m}$$, and for $$\psi_2$$, the energy $$E_2$$ is $$E_2=\hbar \omega_2 = \frac{\hbar^2 (k^2+\Delta k)^2}{2m}$$.

Note that $$\psi_1(x,0)$$ and $$\psi_2(x,0)$$ (the two initial wavefunctions) are only different by an overall spatially-dependent phase $$$$\psi_2(x,0) = e^{i \Delta k x}\psi_1(x,0)$$$$ However they are clearly different states. For example, the energies are different, as noticed above. Furthermore the momenta are different $$\begin{eqnarray} p \psi_1 &=& -i \hbar \frac{d\psi_1}{dx} = \hbar k \psi_1 \\ p \psi_2 &=& -i \hbar \frac{d\psi_2}{dx} = \hbar (k+\Delta k) \psi_2 \end{eqnarray}$$ In particular, the momenta differ by $$\hbar \Delta k$$.

In other words: knowing initial probability distribution for the position does not fully determine the state! One needs to know the probability amplitude distribution$$^\star$$, which contains more information.

$$\star$$ To be technically correct I should say you need to know the probability amplitudes as a function of position, or the probability amplitudes as a function of momentum, or in general, you need to know the inner product of the state with every member a complete basis. I used the words "probability amplitude distribution" to contrast linguistically with "probability distribution", but please don't take this to mean that the there are any implied mathematical properties beyond defining the probability amplitude at every point in position space.

• Thank you for your answer. It made the concept clearer to me. Since the phases are not physically observable, how do we know the initial probability amplitude distribution? Perhaps, we start with an observation which imples that the wavefunction collapses to an eigenstate of that observable. Then we use that eigenstate as the initial condition for the Schrodinger equation? Mar 11, 2021 at 19:34
• @user713320 In fact the phases are physically observable! Now, the overall phase (1 number) is not observable. But the spatially dependent part of the phase is observable. If you transform the wavefuncton from the position basis to the momentum basis, you will find that the probability distribution for momentum depends on the phase of the wavefunction in the position basis. You may find this question relevant: physics.stackexchange.com/questions/80434/… Mar 11, 2021 at 19:40
• Why do you call it "probability amplitude distribution"? It's not a distribution... Mar 12, 2021 at 9:09
• @user21820 I added a footnote. Mar 12, 2021 at 13:31

The initial wave functions with different phases at different points will in general be associated with different physical states, and thus can lead to different measurement outcomes. The key here is that only the global phase of a statevector doesn't modify the physical state itself, but a local phase leads to a different superposition of states.

This can be seen in a simpler system of qubits: the state $$\frac{1}{\sqrt{2}}(|0\rangle + |1\rangle)$$ is different (even orthogonal to) $$\frac{1}{\sqrt{2}}(|0\rangle - |1\rangle)$$, even though we only changed a local phase on the $$|1 \rangle$$ state.

The case with wave functions in space is analogous: the wave function $$\psi(x)$$ is but a coefficient of expansion of the state $$|\psi \rangle$$ in the position basis $$|x \rangle$$, such that $$\psi(x) = \langle x | \psi \rangle \quad \Leftrightarrow \quad |\psi \rangle = \int_{\mathbb{R}} \psi(x) | x \rangle \ dx$$. Thus, adding a position-dependent phase factor $$\psi(x) \mapsto e^{i \phi(x)} \psi(x)$$ is the same as changing the local phases $$|\psi \rangle \mapsto \int_{\mathbb{R}} \psi(x) e^{i \phi(x)} |x\rangle \ dx.$$

As @Andrew's answer demonstrates, the evolution of these states differing by local phases is also different and can produce different position measurement outcomes after some time evolution.

• Thank you for the examples and the explanation! Mar 11, 2021 at 19:35