# Quantum mechanics basics [closed]

Hello. I'm thinking about where does the basic quantum mechanics things comes from. I mean the forms of operators and a Shroedinger equation. The more intuitive explanation is better.

To get forms of operators and Shroedinger eq., we can start from assumption that in our representation square of absolute value of wavefunction is probability density. Then it became clear that coordinate operator is just a multiplying by variable (obvious, looking for example to mean value expression). Next, Shroedinger equation anyway should be of the form $\frac{\partial \phi}{\partial t} = A \phi$ with some A. To see that $A$ is actually Hamiltonian multiplied by something, we can see that $\frac{\partial \phi}{\partial t}$ is an enegry multiplied by something. I have the only idea to explain this: assume additionally that in our representation defined momenta states are plane waves. Considering them, $i\frac{\partial \phi}{\partial t} = E$ is visible from De Broglie relation $E=\omega$ for plane waves (as well as $p=\frac{\partial \phi}{\partial x}$ visible from $p=k$). Then, the only thing i want explained deeper is De Broglie relations. Finally, now i also think that i want some explanation of introducing wavefunction as complex-valued.

Or maybe there is an other way? I guess, it would be more general to start from commutation relations, but i'm afraid it would be hard and abstract. But i appreciate if you try to explain where are they come from.

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## closed as off topic by Douglas Zare, Andreas Blass, Henry Cohn, Jon Bannon, Chris GerigSep 9 '12 at 17:51

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I think you should try theoretical physics stackexchange. –  Jon Bannon Sep 9 '12 at 13:02
@Jon Bannon I think it is right way to ask on both forums and choose the answer which is more "intuitive" for OP, since it is often happens that physicists and mathematicians see the same subject in different ways. Any way I think that here are many people which might give there opinion on the question, so I do not think it reasonable to close. –  Alexander Chervov Sep 9 '12 at 13:09
There are some "related" questions: mathoverflow.net/questions/102313/… mathoverflow.net/questions/102415/… mathoverflow.net/questions/6200 To Ashley you can find more related questions if you press on "tags" under yours questions. –  Alexander Chervov Sep 9 '12 at 13:22
@Alexander: I agree. I suggested the physics site because of Ashley's response to your answer. –  Jon Bannon Sep 9 '12 at 13:50
Re the de Broglie relation, there are a lot of different ways of getting at this, but one is to recognize that according to relativity, both $(t,\mathbb{x})$ and $(E,\mathbb{p})$ transform as vectors. If the E operator is going to be $\partial/\partial t$, then the $\mathbb{p}$ operator had better be $\nabla$. Note that this is more general than the Schrodinger equation, which is nonrelativistic. –  Ben Crowell Sep 9 '12 at 15:27

Probably the explanation via Heisenberg picture is more intuitive (at least for mathematically minded person like me).

Equations of motion in classical mechanics can be described as Hamilton form as follows:

d\dt f = {H, f}

(here { } are Poisson brackets and these equations reduces to standard Hamilton equations of motions, if you take standard phase space R^2n p_i, q_i and standard Poisson bracket these equation will give d/dt p = -dH/dt ; d/dt q = dH/dt . However they make sense on arbitrary Poisson manifold.)

Quantization in Heisenberg picture turns these equations into

d/dt f = [H, f]

(I have omitted (i/h). )

That is more or less all, modula we need to explain what "[H, f]" is. It is commutator in non-commutative algebra which is deformation quantization of the Poisson algebra of classical observables. It should probably be pointed out here why deformation quantization might be considered as intuitively "clear". To explain this let us go in opposite direction: consider non-commutative algebras depending on parameter "h" such that for "h=0" algebra becomes commutative. The point is that Poisson bracket naturally arise in this step: define {f,g} = lim {h->0} (fg-gf) / h . Exercise - to check that such defined Poisson bracket satisfies the Jacobi identity (it follows from the associativity).

So the moral is that Poisson bracket is "shadow of non-commutativity", more precisely first order of the noncommutative product. The deformation quantization task is not construct the non-commutative algebra from the first this first order "shadow", it has been solved in certain generality by Kontsevich. Well, I am not sure how intuitive these reasons are, at least they are so for me.

Another important remark is to relate Heisenber picture to the Shrodinger one. This is quite easy linear algebra. So Heisenberg picture is evolution of "operators (=matrices)" "f" according to equation d/dt f = [H,f]. The linear statement is the following: such evolution on matrices equivalent to Schrodinger like evolution on VECTORS: d/dt v = H v. You can make this "equivalent" in precise statement in various way e.g. lemma: consider f(t) which satisfy the "Heisenberg" equation d/dt f=[H,f], then vector v(t)= f(t)v_0 will satisfy the "Schrodinger" equation d/dt v =Hv, for any vector v_0.

Another important remark is Stone von Neumann uniqueness theorem. It explains why usually operator corresponding to "p" is d/dq and corresponding to "q" is multiplication by "q".
Theorem says (up to details) that if you consider the algebra [p,q]=1 then it has the unique irreducible representation in the Hilbert space. The meaning of this theorem is the following: you can choose ANY other representation of operators corresponding to "p" and "q" ( not just "p"-> d/dq, q-> mult_q), but ALL these representation will be equivalent. So choose whatever you want and do not care.

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Thanks, that would be great if i could make that to look more intuitive for me. Currently, it's not. –  Ashley Sep 9 '12 at 12:03
I tried to extend the answer in more details. I was in a hurry to post the first draft version of the answer since I see votes to close yours question - if it will be close, no one can add answer, although it still be possible to edit already given answer :) –  Alexander Chervov Sep 9 '12 at 12:55
Classical evolution from 0 to $t$ is given by symplectomorphisms of the phase space. For $t$ infinitesimal you get a symplectic vector field. When the vector field is actually Hamiltonian, its generating function is called energy/Hamiltonian. This is what the Hamilton equations tell you. Applying the art of quantization, you see that quantum evolution is given by an automorphism of the Hilbert space (evolution operator). Infinitesimal evolution is given by a certain endomorphism, which is now called the quantum Hamiltonian. –  Pavel Safronov Sep 9 '12 at 15:55

I'll try to help with intuition on why wavefunctions are complex-valued.

First, it's not actually true that they have to be complex. When you quantize electromagnetic waves, the wavefunction is simply the electric and magnetic fields, which are real. The correct statement is that a spin-1/2 particle's wavefunction has to be complex.

As a simple example of why this is, consider the case of two planar sine waves that are moving in antiparallel directions and then merge and superpose.

If the wavefunction is a real scalar, then at a time when the two superposed waves are 180 degrees out of phase, their sum is identically zero. This violates conservation of energy and, perhaps more importantly, conservation of probability.

If the wavefunction is a complex scalar, then one can prove that solutions of the Schrodinger equation always conserve probability. This is easy to check in an example like the superposition of $e^{i(kx+\omega t)}$ with $e^{i(-kx+\omega t)}$.

To see that this argument doesn't prove that wavefunctions are always complex-valued, consider electromagnetic waves in the same situation of superposition of antiparallel plane waves. Because there is a right-handed relationship among $\mathbf{E}$, $\mathbf{B}$, and $\mathbf{k}$, you can't make both $\mathbf{E}$ and $\mathbf{B}$ cancel. For example, you could choose the polarization so that $\mathbf{E}$ would cancel, but then $\mathbf{B}$ wouldn't.

To see that the fundamental issue is conservation of probability, so that this is really something specific to quantum mechanics rather than classical physics, consider the case of sound waves, which can be represented as real scalar functions $f$ measuring the pressure. Rerunning the same argument about superposing antiparallel plane waves, we find that it's possible for $f$ to cancel, but that's OK, because $f$ doesn't have a probability interpretation. We also still have conservation of energy, because the energy depends not just on $f$ (potential) but also on $\partial f/\partial t$ (kinetic), so $f$ can vanish without making the energy vanish.

Complex numbers come up in a lot of places in quantum mechanics, not just in wavefunctions, and it's not always obvious when they're just a notational convenience. For example, Pauli basically reinvented the quaternions in 1924. His spin matrices $\sigma_1$, $\sigma_2$, and $\sigma_3$ are equivalent to the quaternions i, j, and k if you multiply them by i.

Operators can be complex-valued, but expectation values are always supposed to be real, since they correspond to measurable quantities.

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I don't completely agree. The Hilbert space of states of a quantum system is always a complex vector space. If one defines a wave function as coordinates of a state in a given basis, then it will be in general complex valued. This has no relation with the properties of the classical system that one quantizes. –  user25309 Sep 9 '12 at 16:56
@unknown (google): I agree with your view from 30,000 feet, but there really is a difference between how the complex number system relates to different fields. Under a gauge transformation, the electromagnetic field (expressed in terms of the 4-potential) transforms as $A_j\rightarrow A_j+\partial_j \phi$, while the wavefunction of an electron does $\Psi\rightarrow e^{i\phi}}\Psi$. One involves a complex phase, the other doesn't. Note also that electromagnetic fields are observable, while $\arg\Psi$ isn't. The complex phase ultimately arises from charge, which electrons have and photons don't. –  Ben Crowell Sep 9 '12 at 17:17
When you quantize the electromagnetic field the wave function is most definitely not the electric and magnetic fields. The fields are operators which act on a complex valued wave function, just as in the non-relativistic Schrodinger equation. You are confusing complex valued fields with the complex valued wave function of quantum mechanics. They are two different things. –  Jeff Harvey Nov 26 '12 at 22:09