I searched about the difference between state vector and basis vector in Quantum mechanics but couldn't find any clear explanation. Can someone please give a simple and clear explanation of this?

  • 2
    $\begingroup$ Do you know what the basis of a vector space is? $\endgroup$ – Javier Oct 2 '18 at 13:20
  • $\begingroup$ @Javier , Yes,not rude, seriously, we need to know if you are familiar with the notion of "basis", because the answer can be really long if we need to explain basis, but else it's short enough. $\endgroup$ – FGSUZ Oct 2 '18 at 13:54
  • $\begingroup$ According to my understanding Basis are just like our x,y z coordinate. And a state vector is a vector that is expressed in these coordinates $\endgroup$ – herry Oct 2 '18 at 13:59

Basis vectors are a special set of vectors that have two properties:

  1. The vectors in the set are linearly independent (meaning you cannot write one vector as the linear combination of other vectors in the set)
  2. Every vector in the vector space can be written as a linear combination of these basis vectors

Basis vectors are widely used in linear algebra and are not unique to quantum mechanics.

When we start talking about state vectors in QM, like $|\psi\rangle$, we can choose to express this state vector in terms of any basis we want. In other words, for a discrete basis:

$$|\psi\rangle=\sum_i c_i|a_i\rangle$$

where $|a_i\rangle$ represents basis vector $i$, and $c_i$ is a coefficient saying "how much of $|a_i\rangle$ is in $|\psi\rangle$

Now it could be that $|\psi\rangle$ is equal to one of our basis vectors, say $|a_j\rangle$, so that $c_i=\delta_{i,j}$ and $$|\psi\rangle=\sum_i \delta_{i,j}|a_i\rangle=|a_j\rangle$$

But this is a unique case. We could even choose express this example in some other basis: $$|\psi\rangle=|a_j\rangle=\sum_id_i|b_i\rangle$$

So to answer the question: basis vectors are just a special set of vectors with the two properties listed above. Each basis vector could be a state vector, if the system is purely in that state, but it does not have to be that way. You can get the entire picture by being more general: state vectors can be expressed as linear combinations of basis vectors of whatever basis we choose to work in. This then covers the case for when our state vector is one of our basis vectors, since this is still the case of a linear combination. The choice of basis is completely subjective though (although some bases are better to work in than others for certain problems).

  • $\begingroup$ I was trying to understand this concept in terms of Heisenberg picture. According to my understanding in this picture operators(position x(t) and momentum p(t)) change with time but basis (x,y,z coordinates) remain fix. but in Shankar's book author has mentioned that in Heisenberg picture basis can rotate like state vectors. So in such basis,the vectors appear to be frozen. $\endgroup$ – herry Oct 2 '18 at 14:10
  • $\begingroup$ Shouldn't basis vectors remain fix in Heisenberg picture. Why thhe author says that vectors remain frozen.According to my understanding he should mention that basis should remain fix but vectors can rotate. So keeping this thing in my mind I posted above question . $\endgroup$ – herry Oct 2 '18 at 14:12
  • 2
    $\begingroup$ @herry In the Heisenberg picture, operators evolve in time, but states don't; for that to be consistent with the Schrödinger picture, it is a natural consequence that the basis vectors also need to evolve in time. When you claim that vectors can rotate but the basis show remain fixed, you are implicitly coming back to Schrödinger picture. $\endgroup$ – Bruno De Souza Leão Oct 2 '18 at 14:26
  • $\begingroup$ @herry What Bruno says is correct. If you have further questions about this I would recommend asking a new question. Your current question does not mention anything about this. $\endgroup$ – Aaron Stevens Oct 2 '18 at 14:27
  • 1
    $\begingroup$ @herry You are implicitly defining the basis vectors as the eigenstates of a particular operator, so in order for them to change and still remain eigenstates of a certain operator, this operator in general will also be changing. But that should happen no matter what eigenbasis you pick, so that's why you should expect all operators to be changing with time (the reasoning usually goes the other way around, but I think that might be a little more illuminating to you) $\endgroup$ – Bruno De Souza Leão Oct 3 '18 at 1:51

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.