Why is Schrodinger's Equation complex?

[Colin_Smith]

Hi

Came across a guy in a cafe today who asked the question, " Why is Schrodinger's equation complex?". No one knew the answer and he didn't provide one either.

What is the reason?

Colin

 

[FlexGunship]

"Why is Schrodinger's equation complex?"

Do you mean: "Why are solutions to Schrondinger's equation expressed in the complex plane?"

Schrodinger's original goal was to express phase as a complex factor in polar form.

 

[Theinvoker]

Could be one of many reasons, but if you assume it to be correct it is not that surprising.

In classical physics laws, people are used to X = Y, and that it will always be the case (in ideal conditions). However, in quantum mechanics you are dealing with probability and the equation must therefore describe this range of possible outcomes. In a world where nothing is certain it is not so surprising a complex equation arises.

Or it just gives a very good description, but is fundamentally flawed. There could be lots of reasons really, you decide ;)

 

[Dr Lots-o'watts]

Whenever you have waves, you have sines and/or cosines. But the math is always more compact if you replace sines and/or cosines with e[i(x)].

So Euler's formula allows to shorthand the math, at the expense of alienating people with the mystical, divine, suspicious, strange, eccentric yet so innocent, symbol "i".

 

[Avodyne]

It's complex because that's what turns out to be necessary for the theory of quantum mechanics to agree with experiment.

Of course, you could take the real and imaginary parts of the Schrodinger equation, and write it as two real equations instead. But it's much more elegant to leave it as a single complex equation.

Note that this is not true of the equations of classical mechanics or electromagnetism; they are written as real equations for real variables, and in general cannot be re-written as half as many complex equations. Of course, for particular solutions, such as plane EM waves, it can be useful to express such a solution as the real part of a complex function. But the equations themselves do not simplify for the most general solutions.

Thus, the appearance of i in the Schrodinger equation is a deep property of nature. It's not just a mathematical trick.

 

[Delta2]

It's complex because that's what turns out to be necessary for the theory of quantum mechanics to agree with experiment.

Of course, you could take the real and imaginary parts of the Schrodinger equation, and write it as two real equations instead. But it's much more elegant to leave it as a single complex equation.

Note that this is not true of the equations of classical mechanics or electromagnetism; they are written as real equations for real variables, and in general cannot be re-written as half as many complex equations. Of course, for particular solutions, such as plane EM waves, it can be useful to express such a solution as the real part of a complex function. But the equations themselves do not simplify for the most general solutions.

Thus, the appearance of i in the Schrodinger equation is a deep property of nature. It's not just a mathematical trick.

I think u can reduce Maxwell's equations to 2 if u use the vector V=E+iB, so that curlV=idV/dt+iJ, divV=rho.

What this deep property of nature might be? That we always have to consider pairs of concepts/quantities to understand universe? Like Electric-Magnetic field, Energy-Matter?

 

[arkajad]

One possible reason, though debatable, is that the volume element of the Clifford algebra (aka geometric algebra) Cl(3,0) has square = -1. Cl(3,0) is mysteriously isomorphic to the algebra of complex 2x2 matrices.

Another interesting observation is that 'i" provides the handy link between Hermitian operators (observables) and anti-Hermitian generators of symmetries (thus Noether's theorem comes in easily). This link is what is missing in possible axiomatic versions of QM based on real Hilbert spaces. So, for instance, the time translation group is U(t)=exp(iHt/hbar) while the energy observable is H. Which makes life so much easier!

It is interesting to observe that 'i' in QM is almost always (I think so) accompanied by hbar. Perhaps there is some mystery there to be yet discovered?

Another observation is that something like the imaginary 'i' appears already in symplectic (Hamiltonian) mechanics.

 

[zonde]

Why is Schrodinger's equation complex?

It is convenient way how to insert phase factor into equations that describe phase dependent interactions.

 

[nnnm4]

The need for complex numbers in QM (or the undeniable convenience of using them at least) can be seen in a simple spin-1/2 system. To express the eigenstates of Sx and Sy in the Sz basis and to have the independent states be orthogonal, one needs complex coefficients.

 

[unusualname]

When Schrödinger first discovered his wave equation he did not think of it as fundamentally complex, he really thought he was describing some real-valued physical wave such as distribution of charge. Einstein and others were rather excited that Heisenberg's bizarre abstract matrix formalism had been usurped by the proper physical description in terms of a real physical wave phenomenon.

The modern complex form of the Schrödinger equation does not even appear in the first four seminal papers published by Schrödinger in 1926.

Their initial hopes were quickly dashed however when Born (helped by Pauli) and others realized the correct interpretation of the wave was as a description of probability amplitudes, initially as scattering probabilities (Born) and then as probabilities for position and other measurable quantities.

The wave equation was describing the distribution of complex probabilities with a phase and a modulus, the phase being crucial for describing interference effects in quantum mechanics.

It has to be complex to correctly describe nature, you can not decouple the equation into two equivalent real-valued equations, you end up simply rewriting two real-valued coupled equations which are identical to the single complex valued equation, which is silly.

Schrödinger accepted this and wrote the complex valued form of the equation in his fifth paper published in 1926 (If I recall correctly from Mehra's historical account of the era)

 

[ArjenDijksman]

Hi

Came across a guy in a cafe today who asked the question, " Why is Schrodinger's equation complex?". No one knew the answer and he didn't provide one either.

What is the reason?

Colin

Fundamentally, the mathematical entity representing a quantum particle (or system) is a state vector (an arrow). The orientation of this arrow evolves whatever the variable that changes (time, position, momentum). This means that this arrow rotates or spins in space. So the term describing the derivative of the vector (= d|psi>) with respect to t or x or p is perpendicular to the state vector. Perpendicularity is expressed as exp(i.pi/2)= i. The second derivative will be opposite to the state vector, the third derivative will contain a factor -i, the fourth derivative will be collinear with the state vector, etc.

 

[PhilDSP]

Their initial hopes were quickly dashed however when Born (helped by Pauli) and others realized the correct interpretation of the wave was as a description of probability amplitudes, initially as scattering probabilities (Born) and then as probabilities for position and other measurable quantities.

It's easy enough to forget that phase is not only an important characteristic of waves but also of revolving/rotating objects that are non-spherical in shape. Such an object will be influenced by a wave and vice versa. But of course we can't expect the object normally to be phase-locked with the wave. That's most likely the reason that a "spin" equation for an electron for instance will not properly model an interacting wave.

 

[arkajad]

Let us take the wave function of n particles. Why do we have 3n space coordinates, but only one phase?

 

[unusualname]

Let us take the wave function of n particles. Why do we have 3n space coordinates, but only one phase?

You don't have one phase, you have a phase (and a modulus) defined at every point of 3n dimensional configuration space since the wave function is C(omplex)-valued

In fact, you may want to have a C^2 valued wave function to model another degree of freedom, then you would have 2 phases (and moduli) at every point.

 

[kexue]

Hi

Came across a guy in a cafe today who asked the question, " Why is Schrodinger's equation complex?". No one knew the answer and he didn't provide one either.

What is the reason?

Colin

It's because time evolution of the wave function or a wave vector must be unitary. That means that when you start with a state whose norm is one, the norm of the state stays one while the state evolves in time. A free wave function will be spreading out with time, but the norm will remain the same. It has to be so, because of propability must be conserved. The imaginary unit i in the Schrödinger takes care of that.

 

[unusualname]

It's because time evolution of the wave function or a wave vector must be unitary. That means that when you start with a state whose norm is one, the norm of the state stays one while the state evolves in time. A free wave function will be spreading out with time, but the norm will remain the same. It has to be so, because of propability must be conserved. The imaginary unit i in the Schrödinger takes care of that.

The standard norm isn't conserved, the modulus squared is conserved, this is a consequence of the Schrödinger Equation not the reason for it being complex.

 

[Dickfore]

Because the scalar field over which the Hilbert space of possible states of the quantum system is defined is the set of complex numbers.

 

[kexue]

The standard norm isn't conserved, the modulus squared is conserved, this is a consequence of the Schrödinger Equation not the reason for it being complex.

If a wavefunction is normalized, it stays normalized for all subsequent times, this a direct consequence of the hermiticity of the hamiltonian, since it generates unitary time evolution.

 

[unusualname]

If a wavefunction is normalized, it stays normalized for all subsequent times, this a direct consequence of the hermiticity of the hamiltonian, since it generates unitary time evolution.

You can obfuscate all you like, the thread title asks why the SE is complex, the answer is that it has to be complex to correctly describe nature. Nature can be described in many fancy alternative schemes, but they all require complex numbers.

The wavefunction stays normalized under schrödinger evolution because |Psi|^2 is conserved, this is taught in introductory courses on QM.

 

[kexue]

Unusualname, |Psi|^2 is the norm (of course integrated over space).


No offense taken, but should not you spend a little bit more time learning physics than "teaching" it on an internet forum?

 

[unusualname]

The norm of a complex number can be defined many ways, the modulus squared is good here because it is conserved.

In fact Born originally thought |Psi| could define the probability, but it can't.

Saying the Schrödinger Equation is complex because it preserves a norm is not very accurate, I can think or lots of real-valued equations that preserve a norm.

 

[Dickfore]

@OP:

If the Hamiltonian of the system is time independent, one can always write the time-dependent wave function as:

<br /> \Psi(\mathbf{r}, t) = \sum_{n}{\psi_{n}(\mathbf{r}) \, \exp\left[-\frac{i}{\hbar} \, E_{n} \, t\right]}, \; E_{n} \in \mathbb{R}<br />

where the (real) energies are the eigenvalues of the time-independent Schroedinger equation:

<br /> \hat{H} \, \psi_{n}(\mathbf{r}) = E_{n} \, \psi_{n}(\mathbf{r})<br />

The fact that the eigenvalues are real stems from the fact that the Hamiltonian \hat{H} = \hat{H}^{\dagger} is a Hermitian operator (which is equivalent to saying that the time-evolution operator \hat{U}(t, t_{0}) = \exp\left[-\frac{i}{\hbar} \, (t - t_{0}) \, \hat{H}_{0} \right] is unitary).

Furthermore, if the Hamiltonian possesses time-reversal symmetry, it is real and one can always choose the wave functions to be real!

 

[ArjenDijksman]

The standard norm isn't conserved, the modulus squared is conserved, this is a consequence of the Schrödinger Equation not the reason for it being complex.

The modulus squared being conserved, the modulus (= norm of the state vector, square root of the modulus squared) is also conserved. An evolving state vector can keep constant norm only if its orientation changes, hence the imaginary i (90°) in the evolution equations when the state vector appears with first, third, fifth, seventh... derivatives. So the conservation of the norm requires the Schrödinger equation to contain complex terms.

 

[Dickfore]

Note that this is not true of the equations of classical mechanics or electromagnetism; they are written as real equations for real variables, and in general cannot be re-written as half as many complex equations.

Well, let's take Maxwell's equations in vacuum:

<br /> \begin{array}{rcl}<br /> \nabla \cdot \mathbf{E} &amp; = &amp; \frac{\rho}{\epsilon_{0}} \\<br /> <br /> \nabla \cdot \mathbf{B} &amp; = &amp; 0 \\<br /> <br /> \nabla \times \mathbf{E} &amp; = &amp; -\frac{\partial \mathbf{B}}{\partial t} \\<br /> <br /> \nabla \times \mathbf{B} &amp; = &amp; \mu_{0} \, \left(\epsilon_{0} \, \frac{\partial \mathbf{E}}{\partial t} + \ \mathbf{J} \right) <br /> \end{array}<br />

These equations are linear, so the equation satisfied by any linear combination of the fields should be linear as well. First of all, from the third Maxwell equation it becomes obvious that:

<br /> \frac{[E]}{\mathrm{L}} = \frac{<b>}{\mathrm{T}} \Rightarrow \frac{[E]}{<b>} = \frac{\mathrm{L}}{\mathrm{T}} = [c]<br /> </b></b>

Then, from the fourth equation:

<br /> \frac{<b>}{\mathrm{L}} = \frac{[\epsilon_{0}] \, [\mu_{0}] \, [E]}{\mathrm{T}} \Rightarrow \frac{[E]}{<b>} = \frac{\mathrm{T}}{\mathrm{L} \, [\epsilon_{0}] \, [\mu_{0}]} = \frac{1}{[c] \, [\epsilon_{0}] \, [\mu_{0}]}<br /> </b></b>

Let us define a speed:

<br /> c = \frac{1}{\sqrt{\epsilon_{0} \, \mu_{0}}}<br />

which, of course, is the speed of light in vacuum. Then, \mathbf{E} and c \, \mathbf{B} have the same dimensionality. Then, let us defined a complex valued field:

<br /> \mathbf{X} \equiv \mathbf{E} + i \, c \, \mathbf{B}<br />

The divergence and curl of this field satisfy:

<br /> \begin{array}{rcl}<br /> \nabla \cdot \mathbf{X} &amp; = &amp; c \, \rho_{X} \\<br /> <br /> - i \, \left(\nabla \times \mathbf{X}\right) &amp; = &amp; \frac{1}{c} \, \frac{\partial \mathbf{X}}{\partial t} + \, \mathbf{J}_{X}<br /> \end{array}<br />

where we had defined the charge and current density for the X-field:

<br /> \begin{array}{rcl}<br /> \rho_{X} &amp; = &amp; Z_{0} \, \rho \\<br /> <br /> \mathbf{J}_{X} &amp; = &amp; Z_{0} \, \mathbf{J}<br /> \end{array}<br />

where we had introduced the so called impedance of free space Z_{0} = \sqrt{\mu_{0}/\epsilon_{0}} \approx 377 \, \mathrm{\Omega}.

So, there, I reduced the number of Maxwell's equations to half it's original number. It is true that although my field X is complex, the charges and currents that create it are real. If we had allowed for possible imaginary components and went back and rewrote the Maxwell equations in terms of the standard E and B fields, we would see that now the second and third Maxwell equations acquire extra source terms. This corresponds to free magnetic monopoles!

 

[arkajad]

On physical grounds complex phase may well come from time inversion. For instance Majorana neutrino can live pretty well in a real Hilbert space - gamma matrices in Majorana's representation can be real, therefore also its wave function can be real.

 

[Delta2]

Nice dickforce, a more accurate derivation of what i originally posted. What we didnt notice is that if we look at the equation
<br /> \begin{array}{rcl}<br /> <br /> - i \, \left(\nabla \times \mathbf{X}\right) &amp; = &amp; \frac{1}{c} \, \frac{\partial \mathbf{X}}{\partial t} + \, \mathbf{J}_{X}<br /> \end{array}<br />

and we rewrite it by multiplying both sides by i and take the case of free space where the current density is zero we have :

<br /> <br /> \nabla \times \mathbf{X}\ &amp; = &amp; i \, \frac{1}{c} \, \frac{\partial \mathbf{X}}{\partial t}<br /> <br />

If we now replace the curl operator with the hamiltonian operator so that \mathbf{H} &amp; = &amp; c \hbar \nabla \times

the equation becomes the schrodinger equation \mathbf{H} \mathbf{X}\ &amp; = &amp; i \, \hbar \frac{\partial \mathbf{X}}{\partial t}

So schrodinger equation seems to be a straightforward generalization of the equation of electromagnetic field, with the hamiltonian operator replacing as a more generic operator the curl operator.

 

[Phrak]

So, there, I reduced the number of Maxwell's equations to half it's original number.

I'm quite surprised to see anyone else here on PF has taken a look at complex Maxell equations.

The number of Maxwell's equations can be reduced to zero where all that is left is the assignment of variables to measurable things, which is already implied in the original equations anyway. However, once you introduce complex fields, how do you expect to make the magnetic monopoles go away to correspond with know physical reality, except by demanding the phase in the complex plane is fixed?

 

[arkajad]

Actually all six Maxwell equations can be rewritten as just one equation - by using the real Clifford algebra, for instance Cl(3,0).

 

[Phrak]

Six?

 

[Delta2]

He probably refers to
\mathbf{F} = \left ( \mathbf{E} + \mathbf{v} \times \mathbf{B} \right ) q

\mathbf{J} = \sigma \mathbf{E}

as the additional 2 equations.

 

[arkajad]

Six?

Six or twelve, six complex, twelve real, twenty two "original" - what's the difference, the same order But it hardly can be less than one.

"http://www.sciforums.com/showthread.php?t=104440""

 

[Phrak]

Six or twelve, six complex, twelve real, twenty two "original" - what's the difference, the same order

No, F_{uv} are simply the electric and magnetic field strengths.

Define G=*F. Then dG=J, and dF=0 are Maxwell's equations. G and F are skew symmetric two-forms. d is the exterior derivative, * is the Hodge duality operator. If, however, F is defined as dA, then dF=0 is a mathematical identity (All exact forms are closed.), and the number of independent tensor equations reduces to one.

But it hardly can be less than one.

You'll have to try harder to find the reduction.

An otherwise arbitrary trice differentiable 4-vector field, A is imposed on a pseudo-Riemann manifold. A is the electromagnetic dual 4-vector potential, dA are the electric and magnetic fields, d*dA are the electric current 4-densities, ddA are the magnetic 4-current densities. There are no equations here.

d*dA expresses Ampere's Law and Gauss' Law, and ddA expresses Gauss' law for magnetism and the Maxwell-Faraday equation under the correct identification of the tensor element derivatives of A.

For all this to hang together a subtle ontological distinction is made. It is tacitly assumed that no equivalence relationship is implied between distinct physical elements such as J==d*dA, but that current and charge density are simply aspects of the vector potential.

Where A is complex, additional terms are introduced for currents, d*dA --> d(*+i)dA, where i=sqrt(-1). The wave equation and charge continuity also change.

 

[arkajad]

Did you forget the Lorentz condition d*A=0?

And another comment: I understand that for you General Relativity has no equations at all (thus less than one), because Einstein's equations are nothing else but the definition of the RHS? Right?

 

[Phrak]

Did you forget the Lorentz condition d*A=0?

Again, with the wild cards. Why don't you tell why you think the Lorentz gauge is relevant. Recall that the expressions I gave above are differential not integral.

 

[Dickfore]

The number of Maxwell's equations can be reduced to zero where all that is left is the assignment of variables to measurable things, which is already implied in the original equations anyway.

Please justify this statement.

However, once you introduce complex fields, how do you expect to make the magnetic monopoles go away to correspond with know physical reality, except by demanding the phase in the complex plane is fixed?

The phase of what?

 

[Phrak]

Please justify this statement.

I think I did that in post #32. If there is anything there that needs clarification, I might supplement. But I think what Delta² brought up in post #26 is more interesting and to the topic at hand.

The phase of what?

The phase of the complex 4-potential. Going to a complex 4-vector potential we can gauge fix the phase of the vector to be real valued over all spacetime. This is a global gauge fixing. Then the complex equations reduce to Maxwell's equations and magnetic charge is fixed to zero, everywhere (and everywhen).

I believe a global phase factor is an unmeasurable quantity. A = A' <-- A e^phi

 

[arkajad]

Again, with the wild cards. Why don't you tell why you think the Lorentz gauge is relevant. Recall that the expressions I gave above are differential not integral.

Lorentz gauge is not necessarily relevant in general. But here we are discussing Maxwell equations in the context of wave equations. Wave equations, similar to Schrodinger's equation, usually describe quantum mechanics of elementary particles. Photon supposedly is a spin 1 particle. Without Lorentz gauge we also have longitudinal spin zero states.

 

[Dickfore]

I think I did that in post #32. If there is anything there that needs clarification, I might supplement. But I think what Delta² brought up in post #26 is more interesting and to the topic at hand.

For all this to hang together a subtle ontological distinction is made. It is tacitly assumed that no equivalence relationship is implied between distinct physical elements such as J==d*dA, but that current and charge density are simply aspects of the vector potential.

I'm afraid you had mistaken the cause and effect. J IS the source for d*dA, not the other way around. All currents are produced by moving charged particles in a unique manner. Consequently, external fields act on these particles with a Lorentz force. Your model fails when you try to write an equation of motion for a charged particle in an external field.

The phase of the complex 4-potential. Going to a complex 4-vector potential we can gauge fix the phase of the vector to be real valued over all spacetime. This is a global gauge fixing. Then the complex equations reduce to Maxwell's equations and magnetic charge is fixed to zero, everywhere (and everywhen).

I believe a global phase factor is an unmeasurable quantity. A = A' <-- A e^phi

If you had looked at the equations more carefully, you would have noticed that the sourceless equations (Gauss' Law and Faraday's Law) now have a non-zero right hand side (due to monopoles). This is why the concept of electrodynamic potentials in the usual sense of the word does not have a straightforward generalization.

 

[Phrak]

I'm afraid you had mistaken the cause and effect. J IS the source for d*dA, not the other way around. All currents are produced by moving charged particles in a unique manner. Consequently, external fields act on these particles with a Lorentz force. Your model fails when you try to write an equation of motion for a charged particle in an external field.

It is convenient to begin with a distribution of charge and work out the fields. This convenience does not imply a physically measurable cause and effect such as "charge causes electromagnetic fields," or that "the electromagnetic fields cause the vector potential." I'm open to proof or motivation to this conjecture.

If you had looked at the equations more carefully, you would have noticed that the sourceless equations (Gauss' Law and Faraday's Law) now have a non-zero right hand side (due to monopoles). This is why the concept of electrodynamic potentials in the usual sense of the word does not have a straightforward generalization.

No, a complex vector potential admits magnetic monopoles.

 

[Dickfore]

It is convenient to begin with a distribution of charge and work out the fields. This convenience does not imply a physically measurable cause and effect such as "charge causes electromagnetic fields," or that "the electromagnetic fields cause the vector potential." I'm open to proof or motivation to this conjecture.

But, it's not the question whether electromagnetic fields cause a vector potential, but whether electromagnetic fields cause charges and currents. If you don't think this stance is absurd, I don't have what to say anymore.

No, a complex vector potential admits magnetic monopoles.

Please show your work.