Why "i" in Schrodinger Equation


by the_pulp
Tags: equation, schrodinger
the_pulp
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#1
Apr3-12, 06:43 AM
P: 133
Schrodinger Equation is "i*h*dphi/dt=H*phi"

That is to say that the change in the state is proportional to a linear tranformation of the actual state (I understood the logic behind that), H is hermitian and that means that its eingenvalues are real (right?). But what is the job of i? It has something to do with unitarity or conservation of probabilities perhaps?

Thanks!
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romsofia
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#2
Apr3-12, 09:40 AM
P: 263
If I remember correctly, I believe the i is so we get more results, and then from there, we only take the real ones, however, don't take this for fact until an expert is able to confirm it.

BUT, the H is NOT a hermitian, it is actually a Hamiltonian operator. I'm not sure if you're familiar with with this operator, but to make a long explanation short: The Hamiltonian is used to describe the total energy of the system. The reason we use it is because it is useful for systems that evolve with time.

Hopefully we have an expert comment on why we have i, because I am also curious.
Ken G
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#3
Apr3-12, 10:01 AM
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H is the Hamiltonian operator, but it is, like all operators in QM, also Hermitian (so has real eigenvalues). The purpose of the "i" is to make a first derivative in time have the right phase relationship with a second derivative in space (we want plane wave solutions, after all). The reason there is only a first derivative in time, rather than the usual second time derivative you see in classical wave equations, is that Psi(x) is supposed to contain all the information about the state to figure out what will happen-- you should not need the first time derivative as a boundary condition on an equation that uses a second time derivative. Just why nature acts this way is not so obvious, maybe someone else will have more insight into that.

the_pulp
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#4
Apr3-12, 10:51 AM
P: 133

Why "i" in Schrodinger Equation


Ok, thanks for your reply. I know what the Hamiltonian is. I am not quite satisfied by the answer "we want plane waves" because... I dont need plane waves in my life!. I mean, why do we want plane waves? because its consecuencies agree with experiencie? I was looking something more fundamental. Something like, conservation of probabilities, or isotropy of space, or relativistic covariance (these are just examples that surely have nothing to do with the answer to my question, but point the direction of the answer I was looking).
I mean, if the answer is we need "i" because of "A" and we need "A" because of its concordance with experience, then "A" does not give me more information (to me in particular) than "i" (because I already know that "i" make things agree with experience).

However thanks all the same!!!!!!
Bill_K
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#5
Apr3-12, 12:02 PM
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The Hamiltonian is Hermitian, meaning ∫ψ*(Hψ) dx = ∫(Hψ)*ψ dx. The time derivative operator does not have this property. In fact it follows from an integration by parts that ∫ψ* dψ/dt dx = - ∫(dψ*/dt)ψ dx, showing it is anti-Hermitian. To make a Hermitian operator from the time derivative operator, you need to multiply it by a factor of i.
Ken G
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#6
Apr3-12, 12:04 PM
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Quote Quote by the_pulp View Post
I mean, why do we want plane waves? because its consecuencies agree with experiencie?
Yes, we need a wave equation that will give us agreement with experiments like the two-slit experiment for the free propagation of particles. The wave equation used in QM must explain things like the de Broglie wavelength relation, which was inferred prior to the creation of the mathematics of QM. As in any theory of dynamics, the mathematics is simply chosen to be what works. Note that theories like relativity are based in essentially philosophical constraints about the objectivity of physics, but relativity is not a theory of dynamics, it is a theory of what is invariant between different observers, i.e., what are the things that observers must agree on. So the philosophical constraint relates to what it means to have an objective description of reality, but just what that description is is entirely based on what works (in that example, the speed of light) in concert with some other theory of dynamics (like Newton's laws).
I was looking something more fundamental. Something like, conservation of probabilities, or isotropy of space, or relativistic covariance (these are just examples that surely have nothing to do with the answer to my question, but point the direction of the answer I was looking).
Those are all constraints that are also necessary to make the math work, but I suspect one could find a different hypothetical mathematical system where plane wave solutions do not appear, yet all those other things on your list are satisfied, and it wouldn't be right.

I could be wrong-- sometimes the behavior obeys a surprisingly few number of constraints. But we know the regular wave equation (say for sound waves) satisfies all those things without the "i", it just uses a second time derivative and you need a constraint on how things are changing in time at the start in order to solve for the future behavior. It turns out that quantum mechanics escapes the need of that additional boundary condition, and that's what I think of as the main reason why it uses only a first time derivative. To me, that's actually pretty fundamental-- it relates to the question of whether or not the state of a system, which determines its future, must include how things are changing in the present, or just how things are at a given moment.

Note also that you can treat velocity as a separate aspect of the reality, rather than connecting it to the rate of change of position with time, and then say that Newton's laws are also first order in time. So there are ways to think of Newton's laws where you only need to know the current state, not any time derivatives, but you have to elevate velocity (or momentum more correctly) to having an existence separate from the rate of change of position. Some ways to frame Newton's laws do just that, so the very issue of whether or not the future requires knowledge of how things are changing in the present depends on how you frame your theory. Thus in the opposite vein, there may also be ways to frame quantum mechanics without the "i", but with a second time derivative instead. So questions like yours often depend on how you frame your theory, but they are very good questions to consider.
The_Duck
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#7
Apr3-12, 12:06 PM
P: 796
One aspect of this is that if H is Hermitian (and we need it to be, so that its eigenvalues are real), then we need the i in order to ensure "unitary evolution," meaning basically conservation of probability. Let me take the very simple Hamiltonian operator H = 5, which is clearly Hermitian. Then if I didn't include the i, the Schrodinger equation is

hbar (d/dt)psi = 5 psi

You can convince yourself (by solving the differential equation) that this leads to the norm of psi growing exponentially with time. But include the i:

i hbar (d/dt) psi = 5 psi

and now you can solve the differential equation again and convince yourself that now the norm of psi is constant in time.
Ken G
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#8
Apr3-12, 12:29 PM
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Yes, that's what I mean by the phase relationship between the d/dt part and the H part of the Schroedinger equation, I just took the special case of a plane wave (V=0). The Duck is taking a more powerful special case-- where psi is an eigenfunction of H. Either way, you see that the "i" insures that the d/dt is 90 degrees out of phase (i.e., perpendicular to, in the complex plane) with psi itself, insuring that the psi vector maintains constant magnitude (preserves probability). The eigenvector example is indeed more powerful to consider, because any psi can be expanded on an orthonormal basis of eigenvectors, and if each amplitude of those components of psi changes 90 degrees out of phase with the amplitude itself, then the norm of psi must not be changing in time. The main point is that we have just a first derivative d/dt, so we need an i to make the amplitude change "perpendicular" (in the complex plane) to the amplitude. If we could use a second time derivative, as other wave equations do, we wouldn't need the i. (By the way, this is closely related to the "quantum Zeno effect", which notes that if the amplitudes change perpendicular to themselves, then to first order in time, none of the probabilities can change. Probability changes are second order in time, so they 'take a little while' to appear-- you need to leave the system alone long enough to get changes, constant observation always prevents any changes in what is being observed.)
unusualname
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#9
Apr3-12, 12:39 PM
P: 661
strictly speaking you could specify an anti-hermitian matrix -iH in place of H and then your schrödinger equation doesn't need to have i in it explicitly.

but you haven't got rid of the i's since the eigenvalues of -iH are now purely imaginary, and psi is in any case complex-valued.

the basic reason why i is required in all formulations of qm (schrödinger, heisenberg, feynman path integral) is that nature is described by probability amplitudes which require a phase and a modulus, so pure real numbers will not do, and any model that claims to use them is in fact still describing a phase and a modulus of probability amplitudes (otherwise it isn't describing nature)
Jolb
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#10
Apr3-12, 12:44 PM
P: 419
I feel that the presence of i in the Schrodinger equation is the most unusual and mysterious part about the equation. This is not a trivial question, but there doesn't appear to be any real answer.

One of the basic reasons why i is in there is that we need our wavefunctions in quantum mechanics to be complex. If we were to insist that solutions to the time-dependent Schrodinger equation are real functions with real time derivatives, the only solutions are constants. We only get nontrivial solutions when we allow complex wavefunctions. If i weren't in Schrodinger's Equation, then we could find nontrivial real solutions with real derivatives.

Another example of why we want wavefunctions to be complex with complex derivatives (which wouldn't be ensured without the i in the Schrodinger equation) comes out of time-independent quantum mechanics. We of course believe that a particle can be described as an eigenket of the position operator, i.e. that its probability amplitude is a dirac delta in coordinate space. In this case, we find that this particle's state in the momentum basis is the fourier transform of the dirac delta--which is either identically 1 (for the dirac delta centered at the origin) or a complex plane wave (for anything not centered at the origin).

Here's what I'm talking about:


Thus, when we represent the position eigenket in momentum space, the position of the localized particle is specified by the (complex) phase of the plane wave that represents it. So, without having a complex component to the momentum-space plane wave, the particle can only be located at the origin!
the_pulp
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#11
Apr3-12, 01:01 PM
P: 133
Ok, so, for a couple of you, "i" is there because it´s what assures conservation of probability. I find really delightful this answer. Sorry Bill K, I know that after all, the formalism has to be related to experiments and no matter how "delightful" it is, if it doesn´t match the experiments, the formalism is wrong. But, when I find that some part of a formalism is there only because it makes it match with experiments, I feel deep in my bones that there is something we dont know (and that feeling grows knowing that we are dealing with an incomplete theory -QM or QFT- that does not explain o lot of things like gravity, dark energy, etc.). However, when there is a way to justify from some previously "known" (whatever that means) commonsense, my bones dont feel that way.
In other words, I think that a great part of understanding is empathy. And The Duck response makes me feel happy and Bill K's makes me feel sad (Nevertheless, after all, it always makes me happy that all of you take part of your time to enlight me!).

Thanks
the_pulp
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#12
Apr3-12, 01:49 PM
P: 133
Jolb and Unusual name, thanks for your answer, but, coherently with what I said before, I dont feel happy about it!
I think that the use of complex numbers is something prior to the appearence of "i" in Schrodinger equation. I have a paper, called "Origin of Complex Quantum Amplitudes and Feynman's Rules" written by Philip Goyal, Kevin H. Knuthz and John Skilling which demonstrates that the rules of probability amplitudes depends on the need of using 2 numbers (and not one as in classical probability theory) to calculate probabilities. And I think (just that, my thinking) that the need of two numbers and not ones comes from the lack of knowledge of not only the state of the system measured but also the state of the apparatus used to measure.
So, I still like The Duck response. The "i" is there in order to assure that the probabilities calculated through probability amplitude scheme (being that scheme something defined previously) sum always 1.

I really recommend the paper I mentioned!

Thanks!!
unusualname
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#13
Apr3-12, 03:36 PM
P: 661
Hi,

you don't need the i explicity to preserve the norm, eg if dpsi/dt = H.psi where

H = [ 0 1 ]
......[-1 0 ]

then this has solution

psi(t) = exp(Ht).psi(0) = [ cos(t) sin(t) ] psi(0)
..................................[-sin(t) cos(t) ]

so |psi(t)| = |psi(0)| for all t

this is because H is anti-hermitian and so exp(Ht) is unitary.

However, as I stated above you still have imaginary numbers implicitly since eigenvalues of H are i and -i , I have just multiplied both sides of the schrodinger equation by 1/i to hide i explicitly.
Dickfore
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#14
Apr3-12, 03:49 PM
P: 3,015
Quote Quote by the_pulp View Post
Schrodinger Equation is "i*h*dphi/dt=H*phi"

That is to say that the change in the state is proportional to a linear tranformation of the actual state (I understood the logic behind that), H is hermitian and that means that its eingenvalues are real (right?). But what is the job of i? It has something to do with unitarity or conservation of probabilities perhaps?

Thanks!
Yes it does. The formal solution to the Schrodinger equation is given by:
[tex]
\psi(t) = U(t, t_0) \, \psi(t_0)
[/tex]
where the time-evolution operator is given by:
[tex]
U(t, t_0) = \exp \left[ -\frac{i}{\hbar} (t - t_0) \, H \right]
[/tex]
According to the Principles of Quantum Mechanics, this operator needs to be unitary (probability-preserving). The imaginary unit 'i' is needed for this, since the Hamiltonian, corrsponding to an observable - energy, is Hermitian.
[tex]
U^\dagger(t, t_0) = \exp \left[ -\frac{-i}{\hbar} (t - t_0) \, H^\dagger \right] = \exp \left[ \frac{i}{\hbar} (t - t_0) \, H \right] = U^{-1}(t, t_0) = U(t_0, t)
[/tex]
Dickfore
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#15
Apr3-12, 03:55 PM
P: 3,015
Quote Quote by the_pulp View Post
I am not quite satisfied by the answer "we want plane waves" because... I dont need plane waves in my life!. I mean, why do we want plane waves?
The equations of QM (Schroedinger equation) are linear. Therefore, they may be solved by the method of Fourier transformations. A complete set of orthonormal functions is provided by plane waves [itex]e^{i k x}[/itex]. But, then, it follows that the time dependence for the expansion coefficients is also harmonic [itex]e^{-i \omega t}[/itex], where ω is a function of k - dispersion relation. The dispersion relation for free particles is:
[tex]
\hbar \omega = \frac{(\hbar \mathbf{k})^2}{2 m}
[/tex]
akhmeteli
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#16
Apr3-12, 09:01 PM
P: 584
Quote Quote by unusualname View Post
the basic reason why i is required in all formulations of qm (schrödinger, heisenberg, feynman path integral) is that nature is described by probability amplitudes which require a phase and a modulus, so pure real numbers will not do, and any model that claims to use them is in fact still describing a phase and a modulus of probability amplitudes (otherwise it isn't describing nature)
Pure real numbers will do, at least in some key cases. Please see http://www.physicsforums.com/showpos...68&postcount=9
Ken G
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#17
Apr4-12, 10:24 AM
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Quote Quote by Bill_K View Post
The Hamiltonian is Hermitian, meaning ∫ψ*(Hψ) dx = ∫(Hψ)*ψ dx. The time derivative operator does not have this property. In fact it follows from an integration by parts that ∫ψ* dψ/dt dx = - ∫(dψ*/dt)ψ dx, showing it is anti-Hermitian. To make a Hermitian operator from the time derivative operator, you need to multiply it by a factor of i.
You can't say it much more clearly and concisely than that!

(One minor nitpick that has nothing to do with the issue-- I don't think the d/dt expression is actually an operator, because it isn't a measurable. It's just an algebraic manipulation with regard to the time parameter, which is somewhat external to the measurables in quantum mechanics. This is a very minor point that is rarely stressed, indeed I have a QM book that states that ihd/dt is the "energy operator". But H is the energy operator, and ihd/dt is just a mathematical relation it must obey, sort of like the RHS of F=ma. I've similarly seen books that say "ma" is the "net force", but I object to that on similar grounds-- it is important that the net force have some separate existence from ma, such that F=ma not be merely a tautological restatement of the same quantity. In the case of F, it is key that it has its own algebra where individual forces add linearly, and for operators, it is important that they have eigenvalue equations where the eigenvalue is a measurable associated with the operator. Then the t dependence results from that eigenvalue, it doesn't determine it. I don't know if there is a good example where this distinction would matter, but it seems like a distinction that is important somehow. All the same, your answer above is dead on, with "operator" replaced by "mathematical expression".)


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