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I think I'm comfortable with the PDE solutions to the Schrodinger equation. But as soon as we start putting these values in a matrix (in dirac notation), I lose my understanding and everything becomes plug-and-chug math magic.

I'm wondering if anyone has an understanding as to why eigenvalues of the eigenstates of a matrix correspond to physical observables. That is, how can we show, using wave mechanics, that the eigenvalues to our eigenfunctions in our PDE can correspond to eigenvalues to our eigenvectors in a matrix? And can we use this understanding to get a better idea of what's going on when we look at the eigenvalues of eigenvectors of a rotation of our matrix?

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  • $\begingroup$ We have an eigenequation $$A |a \rangle = a | a \rangle$$ in abstract vector space. When we use a discrete basis, it becomes a matrix equation $$ \langle n | A |a \rangle = a \langle n | a \rangle$$. When we use a continuous basis, $$ \langle x | A |a \rangle = a \langle x | a \rangle$$it becomes a differential equation. $\endgroup$ – user26143 Oct 14 '14 at 6:23
  • $\begingroup$ "I'm wondering if anyone has an understanding as to why eigenvalues of the eigenstates of a matrix correspond to physical observables." I think you mean the Born rule. It's an independent axiom of the theory which, to this date, has not been invalidated experimentally. Strictly speaking it is only valid in this form in flat spacetime without gravity (even though one can solve for problems with a weak gravitational potential in an ad-hoc way). One can not derive the Born rule from the rest of standard quantum mechanics. $\endgroup$ – CuriousOne Oct 14 '14 at 8:47
  • $\begingroup$ Like most of us you've been taught QM the wrong way around. Forget about Schrodinger's PDE and everything what you've learned so far about QM. Seriously. This is just one special basis and provides little insight compared to the full theory. It should be something discussed after one is acquainted with the basic ideas of QM. Just take Sakurai and start reading. Trust me you'll be happy you learned it the right way around later. $\endgroup$ – Bubble Oct 14 '14 at 11:27
  • $\begingroup$ @CuriousOne I did not realize this axiom had a name. Thank you for that. So I think my question could be rephrased as "How does one go from the Born rule in dirac notation to the Born rule involved in statements in wave mechanics" $\endgroup$ – Steven Sagona Oct 15 '14 at 0:12
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    $\begingroup$ @StevenSagona: Dirac notation is basically nothing else than a way for physicists to write an inner product on vector spaces (that's what a Hilbert space basically is... a vector space with an infinite number of dimension). You have done EXACTLY the same thing in high school (or the first year of college latest) in a finite number of dimensions (n=2 or 3) when you were calculating the length of a vector and the angles between vectors using a scalar product. An operator in a bra-ket is nothing but linear algebra with matrices. $\endgroup$ – CuriousOne Oct 15 '14 at 0:22
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What you are asking is called, in mathematical terms, spectral theorem. I don't know how much you are interested in details, but any self-adjoint operator $A$ (linear partial differential operator on a Hilbert space) can be written as $$A=\int_{\sigma(A)}\lambda dP_\lambda\;,$$ where $\sigma(A)$ is the spectrum of $A$ and $dP_\lambda$ the spectral measure (projection valued measure). If the operator has purely discrete spectrum (with finite multiplicities), i.e. the operator is either compact or with compact resolvent, this reduces to the maybe more familiar form: $$A=\sum_i \lambda_i \lvert \phi_{\lambda_i}\rangle\langle\phi_{\lambda_i}\rvert$$ where $\lambda_i$ are the eigenvalues (repeated if they have multiplicity $>1$), and $\phi_{\lambda_i}$ the corresponding eigenvector.

As you see, there is a natural identification between an operator and its eigenvalues/eigenvectors; and the Dirac notation $\langle\psi \rvert A\lvert\phi\rangle$ is just a way of writing the scalar product between $A\phi$ and $\psi$. Using the decomposition above, we obtain $$\langle\psi \rvert A\lvert\phi\rangle=\sum_i \lambda_i \langle\psi,\phi_{\lambda_i}\rangle\langle\phi_{\lambda_i},\phi\rangle$$ Keep in mind that we can formally construct the "matrix associated to $A$" to be the (infinite) matrix with elements $M_{ij}=\langle e_i \rvert A\lvert e_j \rangle$, with $\{e_i\}_{i\in\mathbb{N}}$ an orthonormal basis of the Hilbert space. This matrix is diagonal if and only if we chose as a basis the eigenvectors of the operator $A$, and the diagonal elements are the corresponding eigenvalues. Obviously this naïve picture fails if the operator has also continuous spectrum.

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  • $\begingroup$ That's the mechanics of linear operators, but it doesn't explain why physical observables arise in this and no other way, which the OP seems to be asking. $\endgroup$ – CuriousOne Oct 14 '14 at 8:49
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    $\begingroup$ @CuriousOne "That is, how can we show, using wave mechanics, that the eigenvalues to our eigenfunctions in our PDE can correspond to eigenvalues to our eigenvectors in a matrix?" So the answer to me seems fit to what he is asking, i.e. the correlation between eigenvalues of a linear partial differential operator and those of a "matrix" in Dirac's notation. It did not seem a question on the foundation of QM measures, but maybe I'm wrong...well, he has answers to both eventual questions so it' s better ;-) $\endgroup$ – yuggib Oct 14 '14 at 9:03
  • $\begingroup$ The OP asked "I'm wondering if anyone has an understanding as to why eigenvalues of the eigenstates of a matrix correspond to physical observables.". That doesn't follow from the linearity of the formalism. It's an independent axiom. Maybe he is asking two questions in one? $\endgroup$ – CuriousOne Oct 14 '14 at 9:05
  • $\begingroup$ @CuriousOne yes, maybe. However your interpretation seems shared (at least by alanf). Let's see what the OP will say. $\endgroup$ – yuggib Oct 14 '14 at 9:18
  • $\begingroup$ I think this the source of my concerns. Do you have a recommendation to read to learn more about spectral theorem? I tried googling it and found really heavy proof focused material. I have somewhat of a math background - so reading through this MIGHT be do-able. But I have a feeling I'll be left with little to no intuition on what's actually going on in the mathematics. $\endgroup$ – Steven Sagona Oct 15 '14 at 0:01
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A measurement is not a primitive in physics. Rather, a measurement is a physical process that takes place according to the same laws of physics as any other physical process. Those same laws apply to the measurement apparatus, to the person doing the measurement and to the records he makes of the measurement. What distinguishes a measurement from any other kind of physical process? One property that is necessary for a particular process to count as a measurement is that it is possible to copy the result. That is, it has to be possible for the result to be present in one system before being copied and in more than one system afterward.

So what sort of operators represent results that can be copied in this way? According to quantum mechanics, systems evolve unitarily. Any unitary operator can be written in the form: $$ U = \Sigma_a e^{i\phi_a}|a\rangle\langle a|, $$ where the $|a\rangle$ form an orthonormal set. To copy a result this operator would have to leave the operator being copied unchanged and the only operators it leaves unchanged are normal operators. The normal operators that represent whether something happens or not are projectors, so the values attached to those projectors, eigenvalues, represent possible measurement outcomes.

For a more detailed discussion see

http://arxiv.org/abs/quant-ph/0703160.

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  • $\begingroup$ I read most of the paper. I can't say I get every step (maybe I'll make a new question to address what I don't get in the paper) Is this saying that the Born rule can be derived by relying on a symmetry argument and non-wavefunction-collapsing axioms of QM? Isn't it a pretty big deal to claim you've solved the measurement problem of QM? Is this paper controversial? $\endgroup$ – Steven Sagona Oct 15 '14 at 0:03

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