Imagine a wavefunction:

$$ \Psi = C_1 \psi_1 + C_2\psi_2 + C_3\psi_3 + C_4\psi_4 $$

where $C_1$, $C_2$, etc are all constants and $\psi_1$, $\psi_2$, etc are eigenfunctions.

If it is given that $\psi_2$ is orthogonal to $\Psi$ does it mean that the probability for the state $\psi_2$ is = 0. If it is so can some one explain this to me in simple terms?


2 Answers 2


What we mean by:

$\psi_2$ is orthogonal to $\Psi$


$$ \langle \psi_2 | \Psi \rangle = 0 $$

If we evaluate $ \langle \psi_2 | \Psi \rangle$ we get:

$$\begin{align} \langle \psi_2 | \Psi \rangle &= \langle \psi_2 | C_1 \psi_1 + C_2\psi_2 + C_3\psi_3 + C_4\psi_4 \rangle \\ &= \langle \psi_2 | C_1\psi_1 \rangle + \langle \psi_2 | C_2\psi_2 \rangle + \langle \psi_2 | C_3\psi_3 \rangle + \langle \psi_2 | C_4\psi_4 \rangle \\ &= C_2 \end{align}$$

where the last result follows because the eigenfunctions are orthonormal.

So if $\langle \psi_2 | \Psi \rangle = 0$ that means $C_2 = 0$ so if we do a measurement on $\Psi$ the probability we'll find it in the state $\psi_2$ is zero.


Do you know what orthogonal means? It means for simple vectors that the dot product is zero. For wavefunctions that the expectation value between the two wave functions is zero.

The construct you have made, if you take the expectation value of psi, and the four psi_i 's are orthogonal to each other the expectation value you will get is C_2. It means that the probability of finding the general Psi in the Psi_2 state is C_2, by construction of your wavefunction, not zero.


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