In Galilean Relativity if there are two objects, the initial positions of the objects, their masses, and the forces acting on the objects is not enough to uniquely determine where the objects will be in relation to each other at a later time. In Galilean Relativity the laws of physics are the same in all intertial reference frames, and in Newtonian Mechanics applying a coordinate transformation to spatial coordinates of the form $x'=x-vt$ doesn't screw up the maths in force equations that describe motion.

In Quantum Mechanics particles don't have a definite position but instead have a wave function that describes the probability of finding the particle at any location. Also velocities and accelerations don't really make sense as concepts in QM they way they do in classical physics. Still given how macroscopic objects that are described with classical physics are made of elementary particles that are described by QM, I think it could be expected that there would be some sort of quantum version of the way that the laws of physics are the same for all inertial reference frames even for systems that don't require SR to be accurately described.

I know one Differential Equation used to describe systems in non relativistic QM is the Schrodinger Equation of the form


and if I assume a particular initial wave function $\Psi(\vec{x},0)$, that is enough to uniquely define the initial laplacian $\nabla^2\Psi(\vec{x},0)$, and so if I also assume a particular initial potential operator $V(\vec{x},0)$, then that is enough to uniquely describe the initial function for the time derivative of the wavefunction $\frac{\partial\Psi(\vec{x},0)}{\partial{t}}$, and then if I assume a certain potential operator across all of space and time $V(\vec{x},t)$, then that is uniquely define the wave through time. This would appear to mean that the Schrodinger Equation as described is non relativistic in the sense that it implicitly assumes a special reference frame used to describe the evolution of the wave function. This becomes more clear when considering the Schrodinger Equation for two free particles


as here it becomes clear that any initial wavefunction $\Psi(\vec{x_1},\vec{x_2},0)$ uniquely defines the two laplacians $\nabla_1^2\Psi(\vec{x_1},\vec{x_2},0)$ and $\nabla_2^2\Psi(\vec{x_1},\vec{x_2},0)$, and with no potential operators to manipulate it becomes clear that if the two masses are given then that is enough to uniquely define the wavefunction $\Psi(\vec{x_1},\vec{x_2},0)$ through time. So it looks like the Schrodinger Equation as given is truly non relativistic in terms of assuming an absolute reference frame. If I have an original wavefunction of the form $\Psi(\vec{x_1},\vec{x_2},t)$, which satisfies the Schrodinger Equation and I try to apply a Galilean Transformation to the wavefunction with respect to $\vec{x_2}$ at each moment in time to produce a new wavefunction $\Psi(\vec{x_1},\vec{x_2}',t)$, then in general the new wavefunction will not satisfy the Schrodinger as given in the original coordinates. Note that $\vec{u}$ is just meant as something to shift coordinates as opposed to a velocity as it would be in classical physics.

I think I've seen a description for how to relatives the Schrodinger Equation, although I can't remember where I found it, and I'm not sure if it's for Galilean Relativity or SR.

A semi naive approach is to start with the wave function $\Psi=e^{i(kx_1-\omega{t_1})}$

then consider that $\frac{\partial\Psi}{\partial{t_1}}=-i{\omega}e^{i(kx_1-\omega{t_1})}$, $\frac{\partial\Psi}{\partial{x_1}}=ike^{i(kx_1-\omega{t_1})}$, $\frac{\partial^2\Psi}{\partial{x_1^2}}=-k^2e^{i(kx_1-\omega{t_1})}$, $i\frac{\partial\Psi}{\partial{t_1}}={\omega}e^{i(kx_1-\omega{t_1})}$, $-\frac{\partial^2\Psi}{\partial{x_1^2}}=k^2e^{i(kx_1-\omega{t_1})}$ so I can say that in this coordinate system the wave function satisfies the Schrodinger Equation if $V=0$.

Now I switch coordinates using the formulas $x_1=x_2-ut_2$, $t_1=t_2$ to produce the new equation $\Psi=e^{i(k(x_2-ut_2)-\omega{t_2})}=e^{i(kx_2-kut_2-\omega{t_2})}$

and so $\frac{\partial\Psi}{\partial{t_2}}=-i(ku+\omega)e^{i(kx_2-kut_2-\omega{t_2})}$, $\frac{\partial^2\Psi}{\partial{x_2^2}}=-k^2e^{i(kx_2-kut_2-\omega{t_2})}$ given that I postulated that the wavefunction satisfies the Schrodinger Equation in the first coordinate system, it doesn't satisfy the plain Schrodinger Equation in the new coordinate system as the spatial second derivative is the same constant multiplied by the wave function as before but the time derivative is a different constant multiplied by the wave function.

Now I set $a(ku+\omega)=\omega$, to get $a=\frac{\omega}{ku+\omega}$ and now I insert $a$ into the Schrodinger Equation in the new coordinate system to get

$$i\frac{\omega}{ku+\omega}\hbar\frac{\partial\Psi}{\partial{t_2}}=-\frac{\hbar^2}{2m}\frac{\partial^2\Psi}{\partial{x_2^2}}$$ which can be rewritten as


I think for this special case this should relativize the Schrodinger Equation using a "Galilean Operator" but I'm not sure if this would work in the general case especially with the derived term being a constant. Also I'm not sure what it would be in the case of there being a non zero potential term or multiple particles in terms of trying to relativize the Schrodinger Equation.

So my question is, is there a differential equation that combines QM with Galilean Relativity, and if so is it a slightly modified Schrodinger Equation or something else? What differential equation combines QM with Galilean Relativity?

  • 2
    $\begingroup$ Related: physics.stackexchange.com/q/56024/2451 More on Galilean covariance of the SE. $\endgroup$
    – Qmechanic
    Commented Nov 23, 2023 at 9:36
  • 1
    $\begingroup$ There's a mistake there - when you perform a Galilean velocity transformation for a two-particle system, you have to transform both particles. You can't just transform one particle as in $\Psi(\vec{x_1},\vec{x_2}',t)$ and expect the equation to retain the same form. The velocity transform refers to the reference frame, and the new reference frame measures both partices with its new coordinates. So you need $\Psi(\vec{x_1}',\vec{x_2}',t)$. If you transform both particles, the equation retains the same form. $\endgroup$
    – Chad K
    Commented Nov 23, 2023 at 9:54

1 Answer 1


Schrödinger's equation for two free particles (i.e. without a potential energy $V$) $$i\hbar\frac{\partial\Psi}{\partial{t}} =-\frac{\hbar^2}{2m_1}\nabla_1^2\Psi-\frac{\hbar^2}{2m_2}\nabla_2^2\Psi$$ is invariant to the Galilei transformation.

More precisely: If the wave function $\Psi(\mathbf{x}_1,\mathbf{x}_2,t)$ is a solution of Schrödinger's equation, then the Galilei-boosted wave function $$\Psi'(\mathbf{x}_1,\mathbf{x}_2,t) = \Psi(\mathbf{x}_1+\mathbf{v}t,\mathbf{x}_2+\mathbf{v}t,t) e^{i(m_1\mathbf{v}\cdot\mathbf{x}_1+m_2\mathbf{v}\cdot\mathbf{x}_2-\frac{1}{2}(m_1+m_2)\mathbf{v}^2t)/\hbar}$$ is a solution as well, as explained (for a single free particle) in the answers to Galilean covariance of the Schrodinger equation.

Note that the Galilei-boosted wave function above is not simply a coordinate-transformed version of the original wave function $\Psi(\mathbf{x}_1,\mathbf{x}_2,t)$ like $$\Psi'(\mathbf{x}_1,\mathbf{x}_2,t) = \Psi(\mathbf{x}_1+\mathbf{v}t,\mathbf{x}_2+\mathbf{v}t,t)$$ but has an additional phase factor to it, which is actually needed to make it a solution of Schrödinger's equation.

  • $\begingroup$ You should mention that you have patched the Galilei transformation to something else that is not even a coordinate transformation. $\endgroup$
    – my2cts
    Commented Nov 23, 2023 at 11:43

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