quantum mechanics - Where does the $i$ come from in the Schrödinger equation?

I am currently trying to follow Leonard Susskind's "Theoretical Minimum" lecture series on quantum mechanics. (I know a bit of linear algebra and calculus, so far it seems definitely enough to follow this course, though I have no university physics education.)

In general, I find these lectures focus a bit too much on the math and not really on the physical motivation behind it, but so be it (if there are other courses aimed at those with reasonable math skills that focus more on physical meaning, let me know!). However, that's only indirectly related to what my question is about.

In Lecture 4, just after the 40-minute mark, Susskind sets out to derive an expression for what he earlier labelled the time-development operator $U$:

$$|\psi(t)\rangle = U(t)|\psi(0)\rangle$$

He starts out as such:

$$U(\epsilon) = I + \epsilon H$$

which makes sense because the change in time will have to be small, i.e. on the order of a small $\epsilon$. However, he then goes ahead and changes this into:

$$U(\epsilon) = I - i\epsilon H$$

which of course is still fine, because we still don't know what $H$ is supposed to be. Now my problem lies with the fact that Susskind then proceeds to derive an expression for $H$ and, subsequently, the Schrödinger equation in which it figures, from the above equation. The $i$ never gets lost and ends up in that equation.

Could we not just as easily have left the $i$ out, or put a 6 or whatever there? Why put $i$ there? I finished the entire lecture hoping Susskind would get back to this, but he never does, unfortunately. (Which is, I guess, another symptom of this course, with which I'm otherwise quite happy, occasionally lacking in physical motivation.)

1. 
   

   For those of you who know this lecture, or similar styles of teaching: am I missing something here?

   
   


2. 
   

   Alternatively, a general answer as to why there is an $i$ in the Hamiltonian and the Schrödinger equation?

   
   

Answer

Let $U$ be an unitary operator. Write $$ U=\mathbb I+\epsilon A $$ for some $\epsilon\in\mathbb C$, and some operator $A$.

Unitarity means $U^\dagger U=\mathbb I$, i.e., $$ U^\dagger U=(\mathbb I+\epsilon^* A^\dagger)(\mathbb I+\epsilon A)=I+\epsilon^*A^\dagger+\epsilon A+\mathcal O(\epsilon^2) $$

Therefore, if $U^\dagger U=\mathbb I$, we must have $$ \epsilon^*A^\dagger+\epsilon A=0 $$

How can we archive this? We can always redefine both $\epsilon$ and $A$ so that $\epsilon$ is real. If you do this, we get $A^\dagger=-A$, i.e., $A$ is anti-hermitian. In principle, this is perfectly valid, but we can do better.

If we choose $\epsilon$ imaginary, we get $A^\dagger=A$. We like this option better, because we like hermitian operators. If $A$ is to be identified with the Hamiltonian, we better have $\epsilon$ imaginary, because otherwise $H$ cannot be hermitian (i.e., observable).

Note that Susskind writes $U=\mathbb I-i\epsilon H$ instead of $U=\mathbb I+i\epsilon H$. This negative sign is just convention, it is what everybody does, but in principle it could be a $+$ sign. This sign doesn't affect the physics (but we must be consistent with our choice). This is similar to certain ODE's in classical mechanics (driven harmonic oscillator, RLC circuits, etc), where we use the ansatz $x(t)=\mathrm e^{-i\omega t}$, with a minus sign for historical reasons.

So, we include the factor $i$ in $U$, and we end up with the Schrödinger equation. Had we not included the $i$, we would have got $$ \frac{\partial\psi}{\partial t}=\nabla^2\psi $$ where I take $\hbar=2m=1$ and $V=0$ to simplify the analysis (this doesn't change the conclusions). Note that this is the heat equation. The general solution of the heat equation is

$$ \psi(x,t)=\int\mathrm dy\ \frac{\psi(y,0)}{\sqrt{4\pi t}}\mathrm e^{-(x-y)^2/4t} $$

No matter what $\psi(y,0)$ is, this solution is non-propagating, non-oscillatory and decays in time. Therefore, the "particles" described by the heat equation don't move, and they slowly disappear! (for example, "stationary" solutions are of the form $\psi(x,t)=\mathrm e^{-Et}\phi(x)$, which goes to zero as $t\to \infty$).

Also, if it were not for the $i$ in Schrödinger equation, $\int\mathrm dx\ |\psi(x,t)|^2$ wouldn't be time independent, so the total probability would change in time, and this makes no sense. Therefore, the $i$ in Schrödinger equation makes the Born interpretation of the wave-function possible!

Some things you might want to check out:

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  Continuous Groups, Lie Groups, and Lie Algebras, for example in http://www.cmth.ph.ic.ac.uk/people/d.vvedensky/groups/Chapter7.pdf

  
  


• 
  

  Wigner's theorem: symmetries are realised through linear/unitary operators, or through antilinear/antiunitary operators.

  
  



• 
  

  Translation operator: in quantum mechanics (and in classical mechanics as well, in a sense), space/time translations are represented through unitary operators, where the generators of such operations are the energy/momentum operators.

  
  


• 
  

  Spectral theorem: Hermitian operators have real eigenvalues.

  
  


• 
  

  The Maximum Principle of the heat equation: if $\psi(x,t)$ solves the heat equation and $t_2>t_1$, then $\psi (t_2,x)\le \psi(t_1,y)\ \forall x,y$, which means $\psi$ "decreases" in time (therefore, probability "is destroyed", or "particles disappear").

  
  


• 
  
  

  Schrödinger versus heat equations