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My confusion started from thinking the quantum superposition principle.

Several website say that the quantum superposition means all state can be represented as infinity superposition of orthogonal states. Also, i see that the Hilbert space is a infinity dimension vector space, so I don't have any question at first. (thanks for telling me it is not correct)

When I try to consider some question about spin, I find that I may have some wrong concepts. for example, in Stern-Gerlach experiment, there are two possible results for the spin of an electron: up or down. A pure state is represented by $$ |\Psi\rangle = \alpha |\uparrow\rangle+\beta |\downarrow\rangle $$ $$ \left | \alpha \right |^{2}+\left | \beta \right |^{2}=1 $$ there are only two superposition state(thanks for telling me it is not correct), which is differ from my understanding of quantum superposition.

At this moment, I recall that an operator is a infinite dimensional matrix, and a state vector is a n*1 matrix. Then, $\widehat{A}|\Psi\rangle$ is a matrix multiplication. For matrix multiplication, the number of columns in first matrix should equal the number of rows in second. if the $ |\Psi\rangle = \alpha |\uparrow\rangle+\beta |\downarrow\rangle$, then its matrix only have two rows, but the operator have infinite columns, therefore $\widehat{A}|\Psi\rangle$ cannot be calculated? I think I must have make something wrong, can anyone help me?

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The meaning of Superposition

Since the electron must have gone through one of the two slits, but we have no way of knowing which one without performing a measurement, the total wavefunction can be written $|\Psi\rangle = \frac{1}{\sqrt{2}}( |\psi_1 \rangle + |\psi_2 \rangle) $, where the $\frac{1}{\sqrt{2}}$ factor is just for normalisation. So the electron is in a superposition of $|\psi_1 \rangle and |\psi_2 \rangle$..

Then the matrix form of the wave function is $$\begin{bmatrix} \psi_1 \\ \psi_2 \end{bmatrix} $$ the operator have infinite columns, so there is no operator can act on the wave function? or... is the matrix multiplication between infinite matrix and finite matrix differ from finite matrix*finite matrix?

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  • $\begingroup$ Every (finite dimensional) linear vector space is also a Hilbert space, I believe, it's just that the completeness property, i.e. every Cauchy sequence of elements has a limit that is also in the space, is trivial. A linear operator can have a finite basis or an infinite basis. An infinite basis can be discrete or continuous or both, so it's really all of the above. $\endgroup$ – CuriousOne Jun 23 '15 at 20:57
  • $\begingroup$ "i see that the Hilbert space is a infinity dimension vector space" : finite dimension is also possible ... "there are only two superposition state" : no, it is a state of the pair, that is a superposition of (pairs) eigenstates. Sorry if I misunderstood. $\endgroup$ – user46925 Jun 23 '15 at 21:11
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It may be confusing that there are more that one Hilbert spaces, but it is something that we also see in describing a particle in space.

Thinking of the energy eigenstaates, in 1D free space you will have a superposition of an uncountable number of momentum eigenstates ($k$). However if you have a infinite well of width $a$, the energy eigenstates are now discrete (still infinite but countably infinite), so the Hilbert Space will be smaller.

The fact that spin lives in a finite dimensional space, tells you that it cannot give you full information about the position of the particle. And in fact you can specify the spin of a particle as $\alpha |\uparrow\rangle + \beta |\downarrow \rangle $ but there will be no information of the position of said particle.

The trouble arising from the action of an operator is solved by recognising that we need to act it on the right part of the wavefunction. It is meaningless to ask for the $\hat p \left[\alpha |\uparrow\rangle + \beta |\downarrow \rangle ]\right ] $, where $\hat p$ is the momentum operator, the same way that is meaningless to ask for the $\hat S_x \psi(\vec x)$, where $\psi(\vec x)$ is the spatial part of the wavefuntion.

In the example that you gave you are thinking of $\hat A$ as an operator acting on the normal wavefunction, instead of an operator that acts on the spin components.

I hope this makes things clearer!

-------- Answer to Edit ----------

The notation you use is not correct, which makes the understanding hard to grasp.

To specify a state in any matrix notation, you first specify a basis on which you work, and the state is defined by the coefficients of the wavefunction on that basis. Usually we choose the basis to be the energy eigenstates.

So in this case you have a basis determined by $|\psi_1\rangle$ and $|\psi_2\rangle$, corresponding to the wave function going on the top and the bottom slit respectively.

(Note that this is not a complete basis, if your electron is actually going backwards you could not describe it by a summation of these two basis states.)

However let's consider a system that can only have these two states. Then the state described by the other answer would be :

$$ |\psi\rangle = \frac{1}{\sqrt{2}}\begin{bmatrix} 1\\1 \end{bmatrix} \Leftrightarrow |\psi\rangle = \frac{1}{\sqrt{2}} (|\psi_1\rangle + |\psi_2\rangle) $$

Since we said we were working in the mentioned basis.

Note however that we can still calculate a momentum (which I mentioned was a infinite acting operator) on this state, since each basis state ($|\psi_i\rangle$) is a spatial wavefuntion, where a momentum is defined.

$$ \langle \hat p \rangle_\psi = \langle \psi | \hat p |\psi\rangle = \\ = \frac{1}{2}\left[(\langle \psi_1| +\langle \psi_2|) \hat p (|\psi_1\rangle + |\psi_2\rangle) \right] $$

And the end terms can be calculated if you know $\psi_1(x)$ and $\psi_2(x)$.

Note however that the fact that this set of basis states is not complete (nor eigenstates) means that as the state evolves, you need further basis states to describe the system, and again, since we are in an unbounded system (no potentials and the particles can go to -infinity no problem), means that the number of basis states to describe any possible system is infinite, which goes back to the fact that the dimensionality of the Hilbert space describing a spatial wavefunction is infinite.

Hope this helps!!!

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  • $\begingroup$ do u mean : "if we choose the right part of the wave function, then the $|\psi>$ will have infinite rows, then the operator can act on it?" $\endgroup$ – fairytale Jun 24 '15 at 6:29
  • $\begingroup$ for example, use $\widehat{A}$(assume it is a operator for spin) to act on the spin components. $\endgroup$ – fairytale Jun 24 '15 at 6:38
  • $\begingroup$ Yes, exactly, you just need to make sure you act with the operator on the part of the wavefunction where it is defined. $\endgroup$ – Francisco Machado Jun 25 '15 at 23:29

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