I have learned that the state vector (ket vector) can have many representations, and the laws of quantum mechanics give one the ability to change between representation.

But there is a subtle point where I found it to be difficult to understand.

For a particle with spin (for example, electron), we can prepare the state to be $|\!\uparrow\,\rangle$ along some axis, but the position shall not be restricted. (i.e. I can have the spin-up electron in any point in the space)

So... It seems that although I know the spin of the particle perfectly, I still can't transform the representation to position representation? (Because there are infinite possible positions).

If I know the ket vector perfectly, in this case $|\!\uparrow\,\rangle$, I should be able to know all other dynamical information about the system, but the above situation seems to be a counterexample.


2 Answers 2


Any quantum system is specified by a some complete set of observables. Once the quantum state vector is specified in terms of one complete set, then as you indicate we can change the representation to any other complete set.

The most basic quantum system is a particle moving in one dimension (with no other internal properties). In this case position (X) is a complete set. momentum (P) is also a complete set. We can convert from a position-representation wavefunction to a momentum-representation wavefunction.

An electron has an additional internal degree of freedom call spin, independent of its position (or momentum). So a complete set of observables for an electron is {Position, Spin} or {Momentum, Spin}. So to completely specify an electron state vector you have to specify both its position and spin. Then you can convert this as a whole to a different representation.

  • $\begingroup$ Thank you very much ! Now I know the meaning of Total wavefunction = wavefunction * spin. $\endgroup$
    – Z. Sun
    Sep 9, 2018 at 16:48

I still can't transform the representation to position representation?

It doesn't make sense to transform between a spin representation of your state and the position representation of your state, since the two representations are for two very different observables. In principle, however, we may write the stateket in the energy eigenbasis, $\{ \lvert\phi_{n}\rangle \}$, for some constants $\{ c_{n}\}$,

$$ \lvert\psi\rangle = \sum_{n} c_{n}\lvert\phi_{n}\rangle $$

$$ \lvert\psi\rangle = \sum_{n}\big(\int_{-\infty}^{\infty} \lvert x\rangle \langle x \rvert dx \big) c_{n}\lvert\phi_{n}\rangle $$

$$ \lvert\psi\rangle = \sum_{n}\big(\int_{-\infty}^{\infty} c_{n}\langle x \rvert \phi_{n}\rangle \lvert x\rangle dx \big) $$

where I have simply inserted the identity operator in parenthesis. In the case of your spin 1/2 system, the energy eigenstates are the spin eigenstates, so what do we do with this inner product, $\langle x \rvert \phi_{n}\rangle$, that is inside the integral?

And we have to confront what the wave function derived from this would inform us about reality, and that is not obvious. The standard way to handle this is instead to have a multi-state, or it is also called a spinor, to describe a system that has intrinsic spin and a spatial position as independent properties:

$$ \lvert\psi\rangle = \lvert \uparrow \rangle \lvert x \rangle $$

where $\lvert x \rangle$ is a position eigenstate. We may use matrix representation to write,

$$ \lvert\psi\rangle = \Big( \begin{matrix}\psi_{1} \\ \psi_{2} \\ \end{matrix} \Big) $$

where $\lvert\psi\rangle$ and $\lvert\psi\rangle$ are the spin and position components, respectively, of this spinor.

As Bruce stated in his answer, the spin of the particle is an internal property of it, so you can choose to (conventiently) talk about the spin without necessarily talking about the particle's position, like here. This is what you do when you write $\lvert\uparrow\rangle $ to specify the state of your system as "spin up" without caring about its position in physical space.

So it is not really a counterexample, since nothing is being violated.


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