# Why are QM wave functions complex?

#### IAN 25

Hi,

Can anyone explain to me why the wave functions in QM must to be complex, other than to make it work when inserted into the Schrodinger equation?

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#### f95toli

Gold Member
Because the both the amplitude AND the phase of the wavefunction matters, and the best way to "encode" both amplitude and phase in QM (and in for example electromagnetics) is to use complex quantities.

If the wavefunction was not complex you could e.g. never get interference.

#### IAN 25

Hi f95toli

Thanks for that and I do understand what you mean. However, it is possible in electromagnetism to describe all the properties of waves, interference included, by superposing sine or cosine waves of different amplitudes - without the need for complex functions. It is true one can represent these EM waves (or any other), by using the relation e^ix = cos x + i sin x but it is for convenience. And the phase can be included in the sine or complex forms. So you do not need a complex wavefunction for interference. There must be an additional reason for complex functions in QM.

#### akhmeteli

Hi,

Can anyone explain to me why the wave functions in QM must to be complex, other than to make it work when inserted into the Schrodinger equation?
Actually, real wave functions can be sufficient, at least in some very general and important cases. For example, the scalar wave function can be made real by a gauge transform (at least locally) in the Klein-Gordon equation in electromagnetic field (Schroedinger, Nature, v.169, p.538(1952)). A real wave function can also be sufficient for the Dirac equation in electromagnetic field: three out of four components of the spinor wave function can be algebraically eliminated from the Dirac equation, and the remaining component can be made real by a gauge transform (at least locally) (please see my article in J. Math. Phys.: http://akhmeteli.org/wp-content/uploads/2011/08/JMAPAQ528082303_1.pdf ).

#### friend

Complex numbers are part of a larger collection of hypercomplex numbers. These include, but not limited to, real numbers, complex numbers, quaternion numbers, and octonion numbers. Each of these types of numbers have their own algebraic properties. They are members of the division algebras. The reals have the algebraic properties of having magnitude, commutativity, and associativity. The complex numbers lose the property of magnitude; which is bigger 1 or i? But the complex numbers are still commutative and associative. Then the quaternions lose the algebraic propertes of magnitude and commutativity, but still have associativity. And the octonions lose magnitude, commutativity, and associativity.

I contend that the only justification for using complex numbers in quantum mechanics is because the wavefunctions have the algebraic property of complex numbers, meaning the wavefunctions lose the ability to be compared in magnitude with each other but are still commutative and associative.

The Schrodinger equation is a formalism that deals with one wavefunction at a time, so no comparison between the magnitude is obvious between many wavefunctions. However, in the formalism of Feynman's path integral, an infinite number of wavefunctions are presented in the construction of the path integral. In this context one can ask which wavefunction is larger than another.

As I understand it, wavefunctions are distributions that are all normalized to one. So how does one compare the magnitude of one distribution with another? In this context it would seem obvious that wavefunctions lose the algebraic property of comparing magnitude and so they can be represented with complex numbers.

#### micromass

Complex numbers are part of a larger collection of hypercomplex numbers. These include, but not limited to, real numbers, complex numbers, quaternion numbers, and octonion numbers. Each of these types of numbers have their own algebraic properties. They are members of the division algebras. The reals have the algebraic properties of having magnitude, commutativity, and associativity. The complex numbers lose the property of magnitude; which is bigger 1 or i? But the complex numbers are still commutative and associative. Then the quaternions lose the algebraic propertes of magnitude and commutativity, but still have associativity. And the octonions lose magnitude, commutativity, and associativity.
Magnitude is really a bad term to use in this case. Both the real and the complex numbers can be normed, so maginute makes sense.
What you mean is that complex numbers are not an ordered field, while the real numbers are.

#### jfy4

I don't have a source for this, but I remember reading somewhere that Dirac said complex numbers were need for QM because we reduced the number of independent components of phase space (position and momentum), by being able to write momentum as the derivative w.r.t. position (and visa versa) and hence lost part of the two degrees of freedom. We can get that back by switching to complex numbers which come with two independent components. But this is just a vague memory that I haven't tried to dissect.

#### cosmik debris

The number of parameters needed to describe a mixed state should be the multiple of the parameters to describe each separate state. The goldilocks answer is that with real numbers the left hand side is too big, with quarternions the left hand side is too small, and with complex numbers...it is just right.

#### diegzumillo

Because the both the amplitude AND the phase of the wavefunction matters, and the best way to "encode" both amplitude and phase in QM (and in for example electromagnetics) is to use complex quantities.

If the wavefunction was not complex you could e.g. never get interference.
That's a very concise way to put it. I was about to drown him in Hilbert spaces :P

#### BacalhauGT

with complex numbers, derive exp(iwt) or something like that is just multiplying by iw (i=sqr(-1)). its easy.

you can write the wave function as a sum of exponential terms (fourier t.) so it works for any wave.

#### IAN 25

Thanks to all who have taken the time to reply. I understand that the complex form of describing a wave makes it easier (sometimes) to manipulate mathematically and that we just take either the Re or Im part when dealing with real waves. However, that doesn't explain why the texts on QM insist that the wavefunction be complex. I am afraid the explanations offered by freind and akhmeteli were in parts, beyond my level of mathematical understanding, which is B.Sc Physics level - not postgraduate. Although I do get what friend was saying about comparing magnitudes but then how can a wavefunction formed from any superposition of other wavefunctions also have unit magnitude?

So, I am still wondering is there an explanation which doesn't require higher level mathematics?

Staff Emeritus
You have a choice. You can either express this as a single complex differential equation, or a coupled set of four real differential equations (the real part, the imaginary part, and the two Cauchy-Riemann conditions). These are mathematically equivalent ways to express the same underlying physics.

As a practical matter, it is much easier to work with the single complex equation.

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#### bhobba

Mentor
Actually that question is a very deep one. One answer is suppose you have some kind of measurement apparatus represented by some observable. Rotate or shift the apparatus by an infinitesimal amount and the requirement that the new states that are now the measurement outcomes are still orthogonal and superposition's should still be superposition's means the underlying vector space is transformed by an infinitesimal unitary operator of the form A = 1 +eU where e is an infinitesimal real number. AA(bar) = 1 = 1 + eU + eU(bar) which implies U(bar) = -U so that U is pure imaginary which of course means the underlying vector space must be complex. Other possibilities such as quaternoins have this property but the simplest are complex numbers.

But probably the deepest answer is that a few reasonable assumptions leads to basically two choices - standard probability theory and QM:
http://arxiv.org/pdf/0911.0695v1.pdf

What QM allows is pure states to continuously change to other pure states and entanglement. If you want either of those then you must use QM and complex numbers. Note that continuity is really the basis of the first augment I gave.

Thanks
Bill

#### IAN 25

Thanks to all who have taken the time to reply. I understand that the complex form of describing a wave makes it easier (sometimes) to manipulate mathematically and that we just take either the Re or Im part when dealing with real waves. However, that doesn't explain why the texts on QM insist that the wavefunction be complex. I am afraid the explanations offered by freind and akhmeteli were in parts, beyond my level of mathematical understanding, which is B.Sc Physics level - not postgraduate. Although I do get what friend was saying about comparing magnitudes but then how can a wavefunction formed from any superposition of other wavefunctions also have unit magnitude?

So, I am still wondering is there an explanation which doesn't require higher level mathematics?
From the latest responses, I guess not!

However, using the generic form of the wave function, psi = e^-i(wt - kx) and linking the operators for Energy and Momentum, one automatically gets the Schrodinger equation, including the ih term, which is complex. So, I think my initial comment is valid. The Schroinger eqation is complex because we assume a complex form for the wavefunction and vice-versa.

#### bhobba

Mentor
So, I am still wondering is there an explanation which doesn't require higher level mathematics?
From the latest responses, I guess not!
I feel for you - I really do. Trouble is if there was an answer that required less modest mathematical background books at a less advanced level would give it. That's part of the reason its a deep and far from trivial issue. For now simply accept there is a reason and when you know a bit more math you can delve into the detail. After all you need a reason to put yourself through that heavy math stuff.

Thanks
Bill

#### DrDu

AA(bar) = 1 = 1 + eU + eU(bar) which implies U(bar) = -U so that U is pure imaginary
I think it implies that U is anti-hermitian. However U may still be representable in a suitable basis as a real matrix, e.g. $U=i\sigma_y$ with $\sigma_y$ being a Pauli matrix.

I rather think the reason we use complex vector spaces is to represent time reversal symmetry without sacrificing energy spectrum to be bounded from below. However I don't remember the details.

#### bhobba

Mentor
I think it implies that U is anti-hermitian. However U may still be representable in a suitable basis as a real matrix, e.g. $U=i\sigma_y$ with $\sigma_y$ being a Pauli matrix.
Anti-Hermitian, Skew-Hermitian, Skew-Adjoint or Pure-Imaginary - same thing.

The terminology comes from being able to break any operator into the form A + iB where A and B are Hermitian. You are correct about the Pauli matrices so I would say suggests rather than proves.

Thanks
Bill

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#### akhmeteli

Thanks to all who have taken the time to reply. I understand that the complex form of describing a wave makes it easier (sometimes) to manipulate mathematically and that we just take either the Re or Im part when dealing with real waves. However, that doesn't explain why the texts on QM insist that the wavefunction be complex. I am afraid the explanations offered by freind and akhmeteli were in parts, beyond my level of mathematical understanding, which is B.Sc Physics level - not postgraduate. Although I do get what friend was saying about comparing magnitudes but then how can a wavefunction formed from any superposition of other wavefunctions also have unit magnitude?

So, I am still wondering is there an explanation which doesn't require higher level mathematics?
Let me try to rephrase what I said.

In quantum theory, if you start with a (scalar) wave function $\psi$ and electromagnetic four-potential $A_\mu$, you can use gauge invariance to get a physically equivalent solution $\varphi$ and $B_\mu$, if $\varphi(x)=\exp(i\theta(x))\psi(x)$ and $B_\mu(x)=A_\mu(x)-\frac{1}{e}\partial_\mu\theta(x)$, where $\theta(x)$ is an arbitrary real function (I am cutting some corners here). So it is pretty obvious that you can always choose $\theta(x)$ in such a way that $\varphi(x)$ is real. So you can do with real wave functions only.

#### psmt

So you can do with real wave functions only.
What about the spin wavefunction for a spin-1/2 particle?

#### jfy4

So it is pretty obvious that you can always choose $\theta(x)$ in such a way that $\varphi(x)$ is real. So you can do with real wave functions only.
Show me. I'll give you the set up: Let $\psi(x)$ be a complex wave function, use this freedom to make
$$\psi(x)\rightarrow \psi'(x) = e^{i\theta}\psi(x)$$
real.

#### copernicus1

One "derivation" of the Schrodinger equation that I think might help you is to start with the de Broglie relation $\lambda=h/p\Rightarrow p=\hbar k$ and the Planck-Einstein relation $E=\hbar\omega$. The de Broglie relation implies that a "particle" with momentum $p$ has some sort of wave associated with it, and the PE relation implies we can express the total energy in terms of the angular frequency. Classically, we are used to describing waves with trig functions, something like $\Psi=A\,\cos(kx-\omega t)$. (I'm assuming we've set the phase to zero.)

Now, you can construct a physically relevant quantum wave equation by starting with the expression for the Hamiltonian: $E=T+V$, where E is the total energy, T is the kinetic energy, and V is the potential energy. In order to get the energy out of an expression like $\Psi=A\,\cos(kx-\omega t)$, you'll need to operate on this with a time-derivative:

$$\hbar\frac{\partial}{\partial t}\left(A\,\cos(kx-\omega t)\right)=\hbar\omega\left(A\sin(kn-\omega t)\right).$$ Notice that we've gotten the energy out front, but in doing so we've changed the cosine function to a sine.

Now, if you want to get kinetic energy out of $\Psi$, you're going to have to take a second derivative (with respect to x) on the other side:

$$\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}\left(A\,\cos(kx-\omega t)\right)=-\left(\hbar^2k^2\right)\left(A\cos(kx-\omega t)\right)=-\frac{p^2}{2m}\Psi.$$ Notice the problem we have here: since we took a first derivative with respect to time to get $E$, and a second derivative with respect to x to get $T$, our trig functions don't match and won't cancel out of the final equation. The solution to this issue is to use the complex form of the oscillating wave function, $\Psi=A\,e^{i(kx-\omega t)}$. Now when we take our derivatives we are always left with the exponential and it will cancel out of the final equation:

\eqalign{-\frac{\hbar^2}{2m}\frac{\partial^2\Psi}{\partial x^2}+V\Psi&=i\hbar\frac{\partial\Psi}{\partial t}\cr \frac{\hbar^2k^2}{2m}A\,e^{i(kx-\omega t)}+V\,A\,e^{i(kx-\omega t)}&=\hbar\omega\,A\,e^{i(kx-\omega t)}\cr \frac{p^2}{2m}+V&=E\cr T+V&=E.}

So replacing the classical wave function with the complex exponential really does help to construct a sensible wave equation obeying the de Broglie and Planck-Einstein relations. Hope this helps!

#### akhmeteli

Show me. I'll give you the set up: Let $\psi(x)$ be a complex wave function, use this freedom to make
$$\psi(x)\rightarrow \psi'(x) = e^{i\theta}\psi(x)$$
real.
$\theta(x)=-\frac{1}{2 i}\ln\left(\frac{\psi(x)}{\psi^*(x)}\right)$

#### jfy4

I don't think that prescription preserves expectation values, in general. Consider:
$$\langle \psi | \hat{p} |\psi \rangle =\int dx \; \langle \psi | x \rangle \langle x | \hat{p} |\psi \rangle = \int dx \; \psi^{*}(x) \left( -i \frac{\partial}{\partial x} \right) \psi(x)$$
then under the transformation
$$\rightarrow \int dx \; \psi'^{*}(x) \left( -i \frac{\partial}{\partial x} \right) \psi'(x) = \int dx \; \psi^{*}(x)e^{-i\theta(x)} \left( -i \frac{\partial}{\partial x} \right) e^{i \theta(x)}\psi(x) \neq \int dx \; \psi^{*}(x) \left( -i \frac{\partial}{\partial x} \right) \psi(x)$$
How is that accounted for?

#### bhobba

Mentor
Hi Guys

One other reason occurred to me. If for simplicity and to preserve superposition's you want translations, rotations, and transformations between different reference frames to be linear you must invoke Wigners Theorem. My understanding is that theorem is only valid in complex spaces - for transformations depending on a continuous real parameter like displacements that theorem requires them to be linear and hence superposition preserving.

Thanks
Bill

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