I was reviewing my quantum mechanics notes, and I was confused on what this expression meant:

$$ |{\psi}\rangle = \sum_{i}|{\omega_i}\rangle\langle{\omega_i}|{\psi}\rangle $$ I understand that it's creating a sum of $n\times n$ matrices, where each matrix contains a single 1 and the rest zeros, and when the product is taken it is pulling the corresponding portion of $\psi$ but I don't understand the purpose of this. It's used in finding measurements in quantum systems, but the only information I seem to find is a statement that it is done but not why it's done, and I want to get a better understanding of the fundamental concepts before moving on. Any clarification or insight would be greatly appreciated!

  • $\begingroup$ I haven't really done a course in quantum but the idea here is to write a function in terms of a weighted sum of other functions. It is similar to how we write a vector as v=3i + j + k, we have those basic vectors which we write the vector in. Similarly we could have written vector in some other peperpendicular basis set $\endgroup$ Oct 25, 2020 at 4:53
  • $\begingroup$ For example if you take smthn like the Fourier series then you are expressing a function in terms a sinusoidal basis ( i think?) $\endgroup$ Oct 25, 2020 at 4:54

2 Answers 2


I was reviewing my quantum mechanics notes, and I was confused about what this expression meant:

First of all, given a vector $|\psi\rangle \in \mathcal{V}^n(\mathcal{C})$ you can always expand it in term of indepedent basis set. Let us suppose an orthonormal complete basis set given by $\{|\phi_i\rangle\}$. $$|\psi\rangle=\sum_ic_i|\phi_i\rangle$$
To find coefficient we can take dot product $$\langle\phi_i|\psi\rangle=c_i$$ putting everything together we get $$|\psi\rangle=\sum_i\langle\phi_i|\psi\rangle|\phi_i\rangle=\sum_i |\phi_i\rangle\langle\phi_i|\psi\rangle =\left(\sum_i |\phi_i\rangle\langle\phi_i| \right)|\psi\rangle$$

That explain the math. One point to note that the object on the right in identity ( a complete set of basis) so whenever we need to do expansion or change of basis etc. we insert a complete set of basis.

Now let's see the physics. To see the implication of the concept we just need to recall the postulate of quantum mechanics. Here I'm writing those which are important for discussion.

If the particle is in a state $|\psi\rangle$, measurement of the variable (corresponding to) $\Omega$ with probability $P(\omega)\propto |\langle\omega|\psi\rangle|^2$.The state of the system will change from $|\psi\rangle$ to $|\omega\rangle$ as a result of the measurement.

Just reading this two times give you reason of above math. One thing to note that basis set is not unique, So one always choose the eigenbasis (correspond to operator of interest) when doing measurement.

Hope this help you. Best wishes!


From a physical point of view, you are always interested in observables, which you describe quantum mechanically by means of some hermitian operator. In fact, a property of Hermitian operators is having real eigenvalues, which correspond to the allowed measurement outcomes for a measurement of the observable of interest. Each of these measurement outcomes is, as an eigenvalue, associated to an eigenstate and the set of all of these forms an orthonormal basis. Physically, these eigenstates describe the post-measurement states for a measurement of the corresponding eigenvalue. In your case, the set of eigenstates $\{|\omega_{i}\rangle\}_{i}$ may be associated to some observable described by the hermitian operator $\Omega$ with eigenvalues $\{\omega_{i}\}_{i}$. Expanding the state $|\psi\rangle$ describing your system as a superposition of these eigenstates is useful to determine the probability for each measurement outcome $\omega_{i}$, indeed given by $|\langle \omega_{i} | \psi \rangle|^{2}$, the absolute square of the coefficients in the expansion you wrote.

  • $\begingroup$ Likewise: the expectation value of $\Omega$, $\langle \psi|\Omega|\psi\rangle$, can actually be evaluated as: $\sum_i\langle \psi|\omega_i\rangle \langle \omega_i|\Omega|\omega_i\rangle \langle \omega_i|\psi\rangle = \sum_i \omega_i |\langle \omega_i|\psi\rangle|^2$. $\endgroup$
    – JEB
    Oct 25, 2020 at 0:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.