What is Schrödinger's Equation?

Harys Dalvi

November 2021

In classical mechanics, arguably the most important equation is Newton's famous $$F=ma$$ It's so simple, it almost feels silly putting it on a separate line in the middle of the screen. But it deserves the spot, because it's so central. With this equation, you can take information about a classical mechanics system and figure out how it will change for all time.

In quantum mechanics, there is a similar equation used to predict the behavior of quantum systems. It is Schrödinger's famous

$$i \hbar \frac{\partial}{\partial t} \Psi(\mathbf{r}, t) = - \frac{\hbar^2}{2m} \nabla^2 \Psi(\mathbf{r}, t) + V(\mathbf{r}, t) \Psi(\mathbf{r}, t)$$

Schrödinger is back row, sixth from the left (or right) in this legendary picture

Putting this on a separate line doesn't feel silly at all. Surprisingly, however, this equation isn't too far off from the principles of classical mechanics and \(F=ma\). Today I am going to show you how to understand this equation a little better using classical physics and lots of math. You will need classical mechanics and calculus, but no quantum background. There are just a few facts about quantum physics you need to accept first:

It's important to note that this does not represent a true derivation: it would be easier to just accept Schrödinger's equation directly rather than accept the above facts and go through all this math. However, hopefully this will give some intuition into Schrödinger's equation and the math behind it.

Classical Basis

You might say “Schrödinger's equation looks nothing like Newton's equation, how can they be analogous?” This is a fair point. In fact, Schrödinger's equation is a little more analogous to conservation of energy. (More precisely, it's based on a Hamiltonian, not a true conservation of energy equation.) The equation for classical conservation of energy, where we'll start, is $$E=K+V$$ Where \(K\) is kinetic energy, \(V\) is potential energy, and \(E\) is total energy. Doesn't this already look a little like Schrödinger's equation? We have one term on the left, and it's the sum of two terms on the right.

From here, we're going to modify this equation step by step until we end up with Schrödinger's equation. First of all, we could write \(K\) in terms of mass \(m\) and velocity \(v\). $$E= \frac{1}{2}mv^2 +V$$ It turns out that in quantum mechanics, the momentum \(p=mv\) will be more helpful to us than the velocity. Luckily, we can write \(K\) in terms of mass and momentum as well.

$$K = \frac{mv^2}{2} = \frac{m^2v^2}{2m} = \frac{(mv)^2}{2m} = \frac{p^2}{2m}$$
$$E= \frac{p^2}{2m} + V$$ Now we are going to transition from classical mechanics to quantum mechanics. To do this, we are going to multiply by the quantum wavefunction \(\Psi\) on both sides, just to get it into our equation. $$\boxed{E \Psi= \frac{p^2}{2m} \Psi + V \Psi}$$ But at this point, our equation is a bad mix of classical and quantum mechanics that doesn't really make sense. We said that we don't deal with exact values of momentum in quantum mechanics, only probability distributions with our wavefunction \(\Psi\). But here we have \(p\) and \(\Psi\) in the same equation, as if we knew exactly what the momentum \(p\) was.

We might not know what \(p\) is, but it turns out we can change \(\frac{p^2}{2m} \Psi\) to something in terms of \(\Psi\) and things we do know, so we can deal with probabilities like we're supposed to.


We don't really know what \(\Psi\) is, since we are keeping it general, but we can write it in general terms. How about this: $$\boxed{\Psi = Ae^{ik_xx} e^{ik_yy} e^{ik_zz} e^{-i \omega t}}$$

Why is only time negative?

This is a tough question, and I couldn't find a satisfactory answer online, but here's one way I found to think about it. From multivariable calculus, we have $$\frac{dx}{dt} = - \frac{\partial \Psi/\partial t}{\partial \Psi/\partial x}$$ If we had \(e^{i \omega t}\) instead of \(e^{-i \omega t}\), this would become $$\frac{dx}{dt} = - \frac{i \omega}{i k_x} = - \frac{\omega}{k_x}$$ Later, we'll find that \(\omega = 2 \pi f\) and \(k_x = 2 \pi p_x/h\). This gives $$\frac{dx}{dt} = \frac{p_x}{m} = - \frac{2 \pi f}{2 \pi p_x/h} = - \frac{hf}{p_x}$$ $$\frac{p_x^2}{m} = -hf$$ But this means frequency is negative, or momentum is imaginary, obviously both making no sense. Therefore, we must have opposite signs for space and time.

This equation doesn't really tell us much about \(\Psi\). We have no idea what the values of these variables are. That's actually a good thing, because we don't have information about \(\Psi\), so we don't want to pretend like we do and make stuff up.

All we are saying is that \(\Psi\) is some number \(A\) times something like \(e^{i \theta}\) a bunch of times. Each \(e^{i \theta}\) term shows that the wave function depends on something in some way: \(x\), \(y\), and \(z\) for position in three dimensions, and \(t\) for time. In other words, our equation translated to English is just saying “the wavefunction depends on space and time in some way.”

There is one more thing the equation is saying. \(e^{i \theta}\) is an oscillating function, so our wavefunction will oscillate like a wave.

How do we know it's a wave of this form? Well, actually we don't. But it turns out that if you have some solutions to the Schrödinger equation, their sum (more precisely, their linear combination) will also be a solution. Also, there is something called the Fourier transform which says that you can write any function as a sum of sine and cosine functions (waves). Putting these two ideas together, if we can derive the Schrödinger equation for a general wave, we can add waves together to make whatever other function we want. This sum will also be a solution since it's the sum of individual solutions.

Kinetic Energy

So we have a wavefunction, and it's a wave. We might be interested in the wavelength (in space) and frequency (in time) of the wave. If we have \(e^{i (a) \theta}\), the “wavelength” would be \(2 \pi/a\), since we make a full circle back to \(e^{0i}=e^{2\pi i}\) once \(\theta\) reaches \(2\pi/a\). That means for \(e^{i (k_x) x}\), our wavelength \(\lambda_x\) is \(2 \pi/k_x\). The same idea applies to \(k_y\) and \(k_z\), for the wavelengths in the \(y\) and \(z\) directions.

But wait, remember the de Broglie equation? $$\lambda = \frac{h}{p}$$ This means that if we have the wavelength for each direction, we can easily find the momentum in that direction. $$p = \frac{h}{\lambda}$$ $$p_x = \frac{hk_x}{2 \pi}, \ p_y = \frac{hk_y}{2 \pi}, \ p_z = \frac{hk_z}{2 \pi}$$ Let's define a new constant, \(\hbar = h/2\pi\), just to clean things up a little. $$\boxed{p_x = \hbar k_x, \ p_y = \hbar k_y, \ p_z = \hbar k_z}$$ There's one problem: we have no idea what all these \(k\) values are. I made them up when we wrote a general equation for \(\Psi\). But something interesting happens if we take the second derivative of \(\Psi\). We find $$\boxed{p_x^2 \Psi = - \hbar^2 \frac{\partial^2 \Psi}{\partial x^2}}$$

Proof $$\Psi = Ae^{ik_xx} e^{ik_yy} e^{ik_zz} e^{-i \omega t}$$
$$\frac{\partial^2 \Psi}{\partial x^2} = (ik_x)^2 Ae^{ik_xx} e^{ik_yy} e^{ik_zz} e^{-i \omega t} = (ik_x)^2 \Psi = -k_x^2 \Psi$$
$$\Psi = - \frac{1}{k_x^2} \frac{\partial^2 \Psi}{\partial x^2}$$ $$p_x = \hbar k_x \implies \frac{1}{k_x} = \frac{\hbar}{p_x}$$ $$\Psi = - \frac{1}{k_x^2} \frac{\partial^2 \Psi}{\partial x^2} = - \frac{\hbar^2}{p_x^2} \frac{\partial^2 \Psi}{\partial x^2}$$ $$\boxed{p_x^2 \Psi = - \hbar^2 \frac{\partial^2 \Psi}{\partial x^2}}$$

For the \(y\) and \(z\) components of momentum, we'll have almost the same equation, just replace \(x\) with the new letter. For the total momentum, we have to add the squares of each component: $$p^2 = p_x^2 + p_y^2 + p_z^2$$ Now we can divide by \(2m\) and multiply by \(\Psi\) on both sides, then plug in the equation for momentum in each component with the second derivatives.

$$\frac{p^2}{2m} \Psi = - \frac{\hbar^2}{2m} \bigg( \frac{\partial^2 \Psi}{\partial x^2} + \frac{\partial^2 \Psi}{\partial y^2} + \frac{\partial^2 \Psi}{\partial z^2} \bigg)$$
If you know multivariable calculus, you might recognize the Laplacian operator, \(\nabla^2\), in here. If not, just consider \(\nabla^2\) to be a special abbreviation for the sum of all these second derivatives. $$\frac{p^2}{2m} \Psi = - \frac{\hbar^2}{2m} \nabla^2 \Psi$$ Remember earlier when we were working with classical mechanics? We said that \(p^2/2m\) was kinetic energy. In our new quantum formula, we are using \(- \frac{\hbar^2}{2m} \nabla^2\) on our wavefunction to get the term corresponding to kinetic energy. Therefore, we say that \(- \frac{\hbar^2}{2m} \nabla^2\) is the operator for kinetic energy. Now let's look back at our old equation, where we just took a classical equation and multiplied by \(\Psi\). $$E \Psi= \frac{p^2}{2m} \Psi + V \Psi$$ Now with our operator, we are one step closer to the Schrödinger equation. $$\boxed{E \Psi= - \frac{\hbar^2}{2m} \nabla^2 \Psi + V \Psi}$$ In fact, what we have now is already a valid form of the Schrödinger equation if we know our value for the energy \(E\). But we can get even more information with more math.

Total Energy

Remember I said it might be interesting to know the wavelength and frequency of the wavefunction? We tried finding the wavelength, and ended up coming much closer to the Schrödinger equation. But let's not forget the frequency! Now let's find the frequency and hope we come even closer. Let's go back to the general wavefunction. $$\Psi = Ae^{ik_xx} e^{ik_yy} e^{ik_zz} e^{-i \omega t}$$ The frequency in time will be based on the \(e^{-i \omega t}\) term. Specifically, the frequency will be \(f=\omega/2\pi\), for similar reasons as the wavelength \(2\pi/k\). Let's see what happens if we take the first derivative with respect to time.

$$\frac{\partial \Psi}{\partial t} = (-i \omega) Ae^{ik_xx} e^{ik_yy} e^{ik_zz} e^{-i \omega t} = (-i \omega) \Psi$$
$$f=\omega/2\pi \implies \frac{\partial \Psi}{\partial t} = (-i \cdot 2\pi f) \Psi$$
But wait, remember \(E=hf\)? That means if we have the frequency, we can easily find the energy. If we plug this into our earlier Schrödinger equation, this gives us $$\boxed{i \hbar \frac{\partial \Psi}{\partial t} = - \frac{\hbar^2}{2m} \nabla^2 \Psi + V \Psi}$$ There, that's the full Schrödinger equation!
Show steps We just had $$\frac{\partial \Psi}{\partial t} = (-i \cdot 2\pi f) \Psi$$ $$f = \frac{E}{h} \implies \frac{\partial \Psi}{\partial t} = (-i \cdot 2\pi E/h) \Psi$$ Solve for \(E \Psi\). $$E \Psi = -\frac{h}{2\pi i} \frac{\partial \Psi}{\partial t} = i \frac{h}{2 \pi} \frac{\partial \Psi}{\partial t}$$ Remember we defined \(\hbar = h/2\pi\). $$E \Psi = i \hbar \frac{\partial \Psi}{\partial t}$$ Let's plug this back into our equation. $$E \Psi= - \frac{\hbar^2}{2m} \nabla^2 \Psi + V \Psi$$ $$\boxed{i \hbar \frac{\partial \Psi}{\partial t} = - \frac{\hbar^2}{2m} \nabla^2 \Psi + V \Psi}$$

When I presented the Schrödinger equation at the beginning of this post, I did one more step to make it a little extra scary. We can consider position, which we can write as a vector \(\mathbf{r}\). The wavefunction and potential energy can then be functions of both position \(\mathbf{r}\) and time \(t\). That gives

$$i \hbar \frac{\partial}{\partial t} \Psi(\mathbf{r}, t) = - \frac{\hbar^2}{2m} \nabla^2 \Psi(\mathbf{r}, t) + V(\mathbf{r}, t) \Psi(\mathbf{r}, t)$$

What is Schrödinger's Equation?

So far, we've shown how to find Schrödinger's equation with classical mechanics, but I haven't really explained what the equation is like I promised in the title. Here are some thoughts about what it all means.

As a summary of what Schrödinger's equation is, you can think of it as a statement of conservation of energy in quantum mechanics. One big difference is that it is a probabilistic equation, since it tells you about the wavefunction. The wavefunction can help you predict what a particle is doing, but you can never be completely sure. In classical mechanics, you can be completely sure, at least if your model is right.

It's interesting to see that we played a lot with the total energy \(E\) and the kinetic energy \(K\) in the transition to quantum mechanics, but the potential energy \(V\) is still just written as \(V\). This makes some sense, since \(V\) really depends on the situation, while \(E\) and \(K\) are properties of the particle itself.

You must be wondering, why is the Schrödinger equation so much more complicated than \(E=K+V\)? Well, it doesn't have to be. It just is that way because it's more explicit about the quantities we need. If you want the simple version, you can write $$E \Psi = \hat{H} \Psi$$ Where \(E\) is energy, and \(\hat{H}\) is called the Hamiltonian operator and equals \(- \frac{\hbar^2}{2m} \nabla^2 + V\). These forms are equivalent, since we earlier proved that \(E \Psi = i \hbar \ \partial \Psi/\partial t\).

It seems like conservation of energy has a simple version in both classical physics and quantum physics. We just did a lot of work with a complicated analog of conservation of energy in quantum phyiscs. Is there an analog to this in classical physics?

In general, for a conservative system, the Hamiltonian represents the sum of kinetic and potential energy. So in classical physics, we have $$E = K+V = H$$ $$E = \frac{p^2}{2m}+V$$ From here, we can't go further without more information, so we could say that this is the classical analog of the complicated Schrödinger equation.

But if we know more about the situation, we can make this more complicated. Maybe we know the potential energy is from gravity, and maybe we know the initial energy was all gravitational potential energy. Then we can have

$$gh_0 = \frac{1}{2}\bigg[\bigg(\frac{dx}{dt}\bigg)^2+ \bigg(\frac{dy}{dt}\bigg)^2+\bigg(\frac{dz}{dt}\bigg)^2\bigg] + gz$$
Proof $$E = \frac{p^2}{2m}+V$$ $$mgh_0 = \frac{1}{2m}p^2 + mgz$$ $$mgh_0 = \frac{1}{2m}(p_x^2+p_y^2+p_z^2) + mgz$$ $$mgh_0 = \frac{1}{2m}(m^2)(v_x^2+v_y^2+v_z^2) + mgz$$ $$mgh_0 = \frac{m}{2}(v_x^2+v_y^2+v_z^2) + mgz$$
$$gh_0 = \frac{1}{2}\bigg[\bigg(\frac{dx}{dt}\bigg)^2+ \bigg(\frac{dy}{dt}\bigg)^2+\bigg(\frac{dz}{dt}\bigg)^2\bigg] + gz$$

This is somewhat complicated, like the more complicated version of the Schrödinger equation. It also has derivatives, which can help us find how the system will change with time. The moral of the story is that given the basic idea of the Hamiltonian \(E=H\), and some information specific to the situation, we can plug in that information to get a more complicated but more useful equation. It's interesting how this works for both quantum and classical physics.

The Schrödinger equation looks scary, and it is a little scary, but it's also meaningful. Just like conservation of energy in the Hamiltonian form can help us tell what will happen in classical mechanics, the Schrödinger equation tells us what will happen to a wavefunction in quantum mechanics.

There's a lot more to this, like how exactly we use wavefunctions, and what a wavefunction is. There are many questions that still don't have answers, like what it means that all this seems probability-based. But considering that this equation is at the heart of our universe (until someone finds a better one that explains quantum gravity or something) it's interesting to know that it's related to classical conservation of energy, which even introductory physics students know about.


  1. Quantum Physics I (B. Zwiebach, MIT)
  2. Hamilton’s Equations of Motion (Jeremy Tatum, University of Victoria)