# Understanding derivation Klein-Gordon equation

The Klein-Gordon Equation is given as follows, using natural units ($$\hbar \to1, c \to 1$$):

$$-\frac{\partial^2 \Psi}{{\partial t}^2} = -\nabla^2 \Psi + m^2 \Psi$$

The way I tried to derive this is as follows, starting from the time-dependent Schrodinger equation:

$$i\frac{\partial \Psi}{\partial t} = \hat{H} \Psi$$,

where $$\hat{H} = \frac{\hat{p}^2}{2m}$$ and $$\hat{p} = i\nabla$$

Now, we are going to make the following substitution:

$$\hat{H} = \frac{\hat{p}^2}{2m} \to \hat{H}^2 = \hat{p}^2+m^2$$

Squaring the time-dependent Schrodinger equation gives:

$$-\frac{\partial^2 \Psi}{{\partial t}^2} = \hat{H}^2 \Psi = (\hat{p}^2 + m^2) \Psi = (-\nabla^2 + m^2)\Psi = -\nabla^2\Psi + m^2\Psi$$

as stated in the beginning.

However, is my derivation valid? For example, I squared the time-dependent Schrodinger equation, but this feels off to me, because I am not squaring the wave function but only the partial derivative and the $$\hat{H}$$ operator. Are there any flaws in this?

(I am not a masters student, so please avoid very complicated symbols).

Because there seems to be some confusion about "squaring the wave function", you need to do no such thing, because you're not "squaring Schrödingers equation". Like hft's answer said, the most straightforward way to produce a quantum version of a classical theory is to consider its Hamiltonian and promote it an operator by formally substituting the observables by operators whose eigenvalues are the physical quantities given by those observables. That is why $$E^2=p^2+m^2$$ becomes $$\hat{H}^2=\hat{p}^2+m^2$$ because applying both sides to an eigenfunction $$|\psi \rangle$$ yields $$\hat{H}\hat{H}|\psi \rangle=E\hat{H}|\psi \rangle=E^2|\psi \rangle=\hat{p}p|\psi \rangle+m^2 |\psi \rangle=p^2 |\psi \rangle+m^2 |\psi \rangle$$ or $$(E^2)|\psi \rangle=(p^2+m^2)|\psi \rangle$$ In short we're interpreting the classical expression as a relation between the eigenvalues of some operators, then we construct them and sub them in.You might have noticed that I wrote $$\hat{H}^2$$ as $$\hat{H}\hat{H}$$ because that is the precise meaning of squaring an operator: you apply it twice.

So now hopefully you see that you don't "square the Schrodinger equation"; you just apply it twice, seeing as if $$\hat{H}|\psi \rangle=i\hbar\,\partial_t |\psi \rangle$$ then $$\hat{H}^2 |\psi \rangle=(i\hbar)^2\partial_t(\partial_t |\psi \rangle)=-\hbar^2\partial_t^2 |\psi \rangle$$

Then subbing this in the classical expression-inspired quantum relationship between the square of the hamiltonian, the momentum and the mass, you get the Klein-Gordon equation $$-\hbar^2\partial_t^2 |\psi \rangle=(\hat{p}^2+m^2)|\psi \rangle$$

The Klein-Gordon Equation is given as follows, using natural units ($$\hbar \to1, c \to 1$$):

$$-\frac{\partial^2 \Psi}{{\partial t}^2} = -\nabla^2 \Psi + m^2 \Psi$$

Yes, this is the Klein-Gordon equation in natural units.

The way I tried to derive this is as follows... However, is my derivation valid?

Basically, yes, it is.

Are there any flaws in this?

There are definitely flaws in using the Klein-Gordon to describe a single relativistic particle, for sure.

But, as far as your "derivation" goes, it is pretty standard.

To say it in words and symbols, one more time.

1. Start from the relativistic equation: $$E^2 = p^2 + m^2$$
2. Assume that the formal replacement $$E\to i\frac{\partial }{\partial t}$$ is valid.
3. Assume that the formal replacement $$\vec p\to -i\frac{\partial }{\partial \vec x}$$ is valid.
4. Blindly make those replacements to arrive at a differential equation: $$-\hbar^2\frac{\partial^2}{\partial t^2}\psi = -\hbar^2 c^2 \nabla^2 \psi + m^2c^4 \psi\;.$$ (Where I have restored the constants $$\hbar$$ and $$c$$ for fun.)

The major flaw with actually trying to use this equation to describe a relativistic particle is that the energy is not guaranteed to be positive.

For example, if $$\psi_0$$ is a plane wave $$\psi_0 = e^{i\vec k\cdot\vec r - i\omega t}$$ then $$E=\hbar \omega$$ and $$\vec p = \hbar \vec k$$, but unfortunately $$E = \pm c\sqrt{\vec p^2 + m^2c^2}$$.

Update:

OP keeps asking in the comments why we don't square the wavefunction. It's not clear to me why one would ever thing to do this... but we certainly don't want to do it, since we want a linear differential equation.

So, to say it a slightly different way in words and symbols, one more time.

1. Start from the relativistic equation: $$E^2 = p^2 + m^2$$
2. Multiple both sides of this equation by $$\psi$$ to get an algebraic equation that is linear in $$\psi$$ (and pretty trivially related to the first): $$E^2\psi = (p^2 + m^2)\psi$$.
3. Assume that the formal replacement $$E\to i\frac{\partial }{\partial t}$$ is valid.
4. Assume that the formal replacement $$\vec p\to -i\frac{\partial }{\partial \vec x}$$ is valid.
5. Blindly make those replacements to arrive at a differential equation: $$-\hbar^2\frac{\partial^2}{\partial t^2}\psi = -\hbar^2 c^2 \nabla^2 \psi + m^2c^4 \psi\;.$$ (Where I have restored the constants $$\hbar$$ and $$c$$ for fun.)
• Thank you very much! I was basically wondering about the squaring process, why does the wave function itself not get squared? Nov 14, 2023 at 21:18
• The things that are being replaced by differential operators ($E$ and $\vec p$) are related to the quantum operators ($\hat H$ and $\hat{\vec p}$) not the quantum wavefunction $\psi$.
– hft
Nov 14, 2023 at 21:21
• I guess another way to point that out is to ask you why you don't "square the wavefunction" in the usual Schrodinger equation. Again, the answer is that there is a "$p^2$" operator that is squared, that $\hat p$ operator can become a differential operator $\nabla$ that gets squared $\nabla^2$, but the wavefunction doesn't get squared. The wave function is still just there for the operators to act on.
– hft
Nov 14, 2023 at 21:22
• Well, no, there's no "mathematical justification," whatever that means... The common sense "justification" is that you want to end up with a linear differential equation, so squaring the function you are operating on would completely ruin that.
– hft
Nov 14, 2023 at 22:30
• Let me say it this way: You have the equation $E^2 = (p^2 + m^2)$. You can multiply both sides of this equation by the same number $\psi$ to get another equation: $E^2\psi = (p^2 + m^2)\psi$. Now you re-interpret the numbers $E$ and $p$ in terms of differential operators to get a linear partial differential equation.
– hft
Nov 14, 2023 at 22:35