Why is Schrodinger's Equation complex?

  • #1
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
 

Answers and Replies

  • #2
FlexGunship
Gold Member
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"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.
 
  • #3
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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 ;)
 
  • #4
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".
 
  • #5
Avodyne
Science Advisor
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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.
 
  • #6
Delta2
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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?
 
  • #7
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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.
 
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  • #8
zonde
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Why is Schrodinger's equation complex?
It is convenient way how to insert phase factor into equations that describe phase dependent interactions.
 
  • #9
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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.
 
  • #10
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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 realised 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)
 
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  • #11
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.
 
  • #12
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Their initial hopes were quickly dashed however when Born (helped by Pauli) and others realised 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.
 
  • #13
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Let us take the wave function of n particles. Why do we have 3n space coordinates, but only one phase?
 
  • #14
662
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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.
 
  • #15
196
1
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.
 
  • #16
662
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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.
 
  • #17
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Because the scalar field over which the Hilbert space of possible states of the quantum system is defined is the set of complex numbers.
 
  • #18
196
1
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.
 
  • #19
662
2
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.
 
  • #20
196
1
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 physic than "teaching" it on an internet forum?
 
  • #21
662
2
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.
 
  • #22
2,967
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@OP:

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

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

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

[tex]
\hat{H} \, \psi_{n}(\mathbf{r}) = E_{n} \, \psi_{n}(\mathbf{r})
[/tex]

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

Furthermore, if the Hamiltonian possesses time-reversal symmetry, it is real and one can always choose the wave functions to be real!
 
  • #23
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.
 
  • #24
2,967
5
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:

[tex]
\begin{array}{rcl}
\nabla \cdot \mathbf{E} & = & \frac{\rho}{\epsilon_{0}} \\

\nabla \cdot \mathbf{B} & = & 0 \\

\nabla \times \mathbf{E} & = & -\frac{\partial \mathbf{B}}{\partial t} \\

\nabla \times \mathbf{B} & = & \mu_{0} \, \left(\epsilon_{0} \, \frac{\partial \mathbf{E}}{\partial t} + \ \mathbf{J} \right)
\end{array}
[/tex]

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:

[tex]
\frac{[E]}{\mathrm{L}} = \frac{}{\mathrm{T}} \Rightarrow \frac{[E]}{} = \frac{\mathrm{L}}{\mathrm{T}} = [c]
[/tex]

Then, from the fourth equation:

[tex]
\frac{}{\mathrm{L}} = \frac{[\epsilon_{0}] \, [\mu_{0}] \, [E]}{\mathrm{T}} \Rightarrow \frac{[E]}{} = \frac{\mathrm{T}}{\mathrm{L} \, [\epsilon_{0}] \, [\mu_{0}]} = \frac{1}{[c] \, [\epsilon_{0}] \, [\mu_{0}]}
[/tex]

Let us define a speed:

[tex]
c = \frac{1}{\sqrt{\epsilon_{0} \, \mu_{0}}}
[/tex]

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

[tex]
\mathbf{X} \equiv \mathbf{E} + i \, c \, \mathbf{B}
[/tex]

The divergence and curl of this field satisfy:

[tex]
\begin{array}{rcl}
\nabla \cdot \mathbf{X} & = & c \, \rho_{X} \\

- i \, \left(\nabla \times \mathbf{X}\right) & = & \frac{1}{c} \, \frac{\partial \mathbf{X}}{\partial t} + \, \mathbf{J}_{X}
\end{array}
[/tex]

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

[tex]
\begin{array}{rcl}
\rho_{X} & = & Z_{0} \, \rho \\

\mathbf{J}_{X} & = & Z_{0} \, \mathbf{J}
\end{array}
[/tex]

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

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!
 
  • #25
1,444
4
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.
 

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