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In position or configuration representation, the Hamiltonian operator, and thus the Schrodinger Equation, is expressed in terms of positions. But the particle never has definite position, what do positions in Schrodinger Equation mean?

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  • $\begingroup$ The positions in the Schrodinger Equation NEVER mean the particle to be of definite position. But I know only some hand-waving reasons lacking of deep insights. $\endgroup$ – Quanhui Liu Aug 4 '18 at 3:00
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Position in Schrödinger equation means position.

Keep in mind the the equation is for the wave function, and you solve for the value of the wave function at various points in space. This really is exactly like the position that appears in an expression for the electric of magnetic field, and should not be at all mysterious.

Now, the relationship between the putative "position of a particle" and the wave-function is—indeed—a more subtle matter, but it has no bearing on the meaning of the position in the equation which is about wave functions and not particles.

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Positions can be thought of as eigenstates of the particle. Consider a system (a particle) which has possible eigenstates $S=|1\rangle, |2\rangle,...$. Now let us say it is in some (non-eigenstate) state $|s\rangle$. If we want to write the state as a function of the eigenstates, we will write the wavefunction $\Psi(i)=\langle i|s\rangle$. The standard wavefunction, i.e. the state as a function of position, is the same thing: we write $\Psi(x)=\langle x|s\rangle$. It's just that the position eigenstates form a dense infinity, instead of being countable or even finite as in usual QM.

It's important to note that no particle will ever be in its position eigenstate, due to the uncertainty principle, so that these are theoretical states. We're just decomposing the state $|s\rangle$ into abstract "position states" by taking the state's projection onto these states, $\langle x|s\rangle$. So the positions in the Schrodinger equation are indexes of these abstract states.

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I guess I'm echoing/rephrasing previous answers, but just because the particle might not have a well defined position doesn't mean that position isn't a thing.

There wavefunction has some value at each point in space which we can then relate to the probability of finding the particle at that (well defined) position

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The particle doesn't have a position, but a one-particle system has a state vector of the form $\int dx\psi(x)|x\rangle$, where $|x\rangle$ owes its label to the eigenequation $\hat{x}|x\rangle=x|x\rangle$ (the operator $\hat{x}$ is nondegenerate). All $x$ does in $\psi(x)$ is show how the integrand varies. The function $\psi$, obtainable by solving the Schrödinger equation, gives the integrand as well as the probability density $|\psi|^2$ for a measurement of $x$.

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A Schrödinger particle may not have a precise position but by Born's rule the probability of finding it at a position $\vec r$ is $|\psi(\vec r) |^2$.

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When you take Schrodinger's equation from the classical wave equation, momentum operator comes from the space derivative, and the energy operator from the time derivative.

You see the position operator in the Hamiltonian, right when you have a potential. But when you are using a position form (Schrodinger wave equation) then the position operator will be a multiplication of the wave function by a real number.

The position operator is used too as momentum shift operator (used as exponetial), just like the momentum operator is the position shift operator, for Galilean boosts.

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Pay attention, don’t confuse the uncertainly principle with something that is related to the wave-function: it is not! The wave function $$\langle x | \psi \rangle = \psi(x)$$ is a complex number, more specifically, is in general a complex function, that satisfies the Schroedinger equation that is a partial differential equation. There is nothing strange up to now, so, the position that appears is just the position, nothing else. The “strange” thing, come from the Heisenberg uncertainly principle that is a statement between OPERATORS that acts on an Hilbert space in which the vector states $|\psi\rangle $ live: as you can see, there is nothing about the wave function in this last part.

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protected by Qmechanic Aug 4 '18 at 16:53

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