This came up while working on a question about measuring the angular momentum of a particle in a superposition of angular momentum eigenstates:

Given that: $$\langle\theta,\phi|\psi\rangle \propto \sqrt{2} \cos(\theta) + \sin(\theta)e^{-i\phi} - \sin(\theta)e^{i\phi}$$

What are the possible results and the corresponding probabilities for measurements of $\hat{L}^2$ and $\hat{L}_z$? $\hat{L}^2$ is simply $2\hbar^2$ as all three terms are eigenstates of $\hat{L}^2$ with eigenvalues $2\hbar^2$

However the three terms are eigenfunctions $\hat{L}_z$ with different eigenvalues, namely $0$, $\hbar$ and $-\hbar$. Now my question is whether I first have to normalise the eigenfunctions and then take the modulus squared of the coefficients to find the probabilites of measuring the corresponding eigenvalue, or whether it is possible to straight away write down: $$p(L_z=0)=\frac{|\sqrt{2}|^2}{|\sqrt{2}|^2+|1|^2+|-1|^2}$$

So basically my question is:

Given a wave function $|\psi\rangle$ and an operator $\hat{A}$, with eigenvalues $\lambda_i$ and non-normalised eigenfunction $|a_i\rangle$, and: $$|\psi\rangle = \sum_i{c_i|a_i\rangle}$$ Is it still true that the probability of obtaining a measurement $\lambda_i$ is given by $p_i=|c_i|^2$?

  • $\begingroup$ You have to normalize the state, otherwise the probabilities of distinct results won't add to 1. When you divided by the norm squared in the angular momentum state example, you intuitively tried to take care of this, but check the actual normalization of the eigenfunctions! The general statement that you made at the end is not true for a non-normalized state. $\endgroup$
    – secavara
    Jan 28, 2018 at 17:11

3 Answers 3


Yes and no. You can just normalize the results with:

$$ \langle \psi | \psi \rangle$$

but you have computed that incorrectly. Remember, the differential solid angle is:

$$ d\Omega = d(\cos{\theta})d\phi, $$

you have used:

$$ d\Omega = d\theta d\phi.$$

I suggest you verify with a table of $l=1$ spherical harmonics.


Given a generic vector $\psi$, the expectation value of an observable $A$ over it is given by

$$\operatorname{EXP}_\psi[A] =\frac{\langle\psi|A|\psi\rangle}{\langle\psi|\psi\rangle}.$$

You can check the normalisation by using the identity operator in place of $A$. Hence you get a well-defined state on the C*-algebra of observables.

The probability of obtaining $\lambda_i$ can be obtained by evaluating the expectation value of the projection $E_i=|a_i\rangle\langle a_i|$, and this shows you that you are missing a "corrective" factor of $\frac1{\langle\psi|\psi\rangle}$. If your vector state is not normalised, then $\langle\psi|\psi\rangle\neq1$ and therefore it must be taken into account.

The point is that, even though you are not forced to use normalised states, sooner or later you will have to normalise anyway, else you'd end up with a total probability that differs from 1.

  • $\begingroup$ not an ideal notation as EXP is usually used for the exponential.... $\endgroup$ Jan 28, 2018 at 17:41
  • $\begingroup$ I have never seen EXP used to denote the exponential function in my life $\endgroup$
    – Phoenix87
    Jan 28, 2018 at 17:43
  • $\begingroup$ I guess that makes us even as I’ve never seen EXP used for expectation value. :) $\endgroup$ Jan 28, 2018 at 17:52

Let me be more specific. Take into account the normalization of the spherical harmonics (link). That angular momentum state should then be written more explicitly as $\psi(\theta,\phi) = C \, 2\sqrt{\frac{2 \pi}{3}} \left( Y_{10} + Y_{11} + Y_{1 \, -1} \right)$, where $C$ is a normalization constant. Since the $Y_{lm}$ form an orthonormal basis, it follows that the right normalization is $\psi(\theta,\phi) = \frac{1}{\sqrt{3}} \left( Y_{10} + Y_{11} + Y_{1 \, -1} \right)$, from which one finds that each state is equally likely.

As you see, normalizing each independent component eigenstate as well as the total wave function becomes relevant when computing probabilities.


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