Is there a space-independent Schrödinger equation?

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1. Mar 11, 2015

Rimmonin

I'm learning about the Schrödinger equation in one of my uni courses, and we've recently gone past how to solve the time-independent version. That got me wondering if there is a space-independent version of the Schrödinger equation and what it could possibly be used for. I know I'm probably talking nonsense to someone who really knows QM :P But I just thought it'd be interesting to contemplate whether space-independence actually makes sense, if there even is such a thing.

2. Mar 11, 2015

atyy

http://quantummechanics.ucsd.edu/ph130a/130_notes/node124.html

It is not called that way, but when you obtain the time-independent Schroedinger equation via separation of variables, you write the full wave function as a product $\psi(x,t) = u(x) T(t)$. The equation for $u(x)$ is the time-independent Schroedinger equation, and the equation for $T(t)$ is clearly space-independent.

In this context, you should bear in mind that this is only an intermediate step in obtaining the full solution to the time-dependent Schroedinger equation, and the time-independent Schroedinger equation is just a mathematical trick.

However, the time-independent Schroedinger equation is physically important in another context, which is when the energy of the system is measured. In the simplest form of measurement, the measurement outcome is an eigenvalue of the measured observable. In this context, the time-independent Schroedinger equation is the eigenvalue equation for the energy observable, and so determines the possible values one can get when energy is measured.

An important thing to keep in mind is that the full time-independent Schroedinger equation is important for how the wave function evolves between measurements, but not when a measurement is made. Between measurements, the time-evolution is reversible because of the time-independent Schroedinger equation. But when a measurement is made, one gets a measurement result from the possibilities determined by the eigenvalue equation, and the wave function collapses.

Last edited: Mar 11, 2015
3. Mar 11, 2015

ddd123

The solutions to the time-independent Schroedinger equation are stationary states, which are states with time-independent probability density. By analogy, the solutions to the equation for T(t) above have a space-independent probability density, which (compatibly with the full time-dependent Schroedinger equation) would be the plane waves.

4. Mar 11, 2015

Rimmonin

Yes, but in separating the variables for TISE we must presume the potential energy U(x) is time-independent... from my understanding this is the mathematical trick and that's why the solutions to TISE give stationary states only (potential energy independent of time). But most cases in reality the potential energy is not time-independent so it should really be something like U(x,t). So what I would really like to ask is what would happen if we chose to solve the equation with U(t) rather than U(x)?

5. Mar 11, 2015

atyy

Last edited by a moderator: May 7, 2017
6. Mar 11, 2015

dextercioby

7. Mar 11, 2015

atyy

But isn't the lattice index conceptually the same as space?

8. Mar 11, 2015

dextercioby

The dynamical variables are only spins (spin matrices). There's no coordinate operator in the Hamiltonian, that's what I meant.

9. Mar 12, 2015

f95toli

It also depends on what you mean by "space".
In many cases you are solving the Schroedinger equation not for real particles, but for electronic systems where the "x" variable does not represent a spatial dimension (i.e, some sort of position) but e.g. phase or charge of the electrons in the circuit. A concrete example would be Josephson junctions where the "x" variable represent the phase of the superconducting condensate.
This can be also be done for large (macroscopic) electronic circuits such as solid state qubits.

10. Mar 14, 2015

samalkhaiat

Schrodinger equation describes the time evolution of a (state) vector $| \Psi \rangle$ in an abstract vector space $\mathcal{H}$ (the Hilbert space) $$i \frac{\partial}{\partial t} | \Psi \rangle = H | \Psi \rangle .$$ Written this way, it is clear that Schrodinger equation is independent of any representation spaces. And, it is exactly this property that makes QM so powerful and universal. To understand what I mean, let $\{ | \alpha \rangle \} \in \mathcal{H}$ be complete orthonormal set of states with the variable $\alpha$ takes on continuous, discrete or continuous and discrete values. Using the completeness property together with the inner-product in $\mathcal{H}$, we can transform the above Schrodinger equation to the following form $$i \frac{\partial}{\partial t} \langle \beta | \Psi \rangle = \sum_{\alpha} \int d \alpha \ \langle \beta | H | \alpha \rangle \langle \alpha | \Psi \rangle , \ \ \ (1)$$ or $$i \frac{\partial}{\partial t} \Psi ( \beta ) = \sum_{\alpha} \int d \alpha \ \langle \beta | H | \alpha \rangle \ \Psi ( \alpha ) , \ \ \ (2)$$ where $\beta$ may or may not be in the set $\{ | \alpha \rangle \}$. I will suppress the time-dependence of the state vector, i.e, you should read $\Psi ( \cdots )$ as $\Psi ( \cdots , t )$.
(i) If we take, in (2), $\beta = x$ and $\alpha = y$ to be coordinates in the position space, we obtain the usual Schrodinger differential equation: $$i \frac{\partial}{\partial t} \Psi ( x ) = \int d^{3} y \ H ( y ) \ \delta^{3} ( x - y ) \ \Psi ( y ) = H ( x ) \Psi ( x ) . \ \ \ (3)$$
(ii) If you take $\beta = p$ and $\alpha = \bar{p}$ to be continuous momentum-space variables, then Schrodinger equation (2) becomes an integral equation (as it should be in momentum space) $$i \frac{\partial}{\partial t} \Psi ( p ) = \int d^{3} \bar{p} \ H ( p , \bar{p} ) \ \Psi ( \bar{p} ) . \ \ \ \ (4)$$
(iii) Now consider $( \beta , \alpha )$ to be discrete energy levels $( E_{n} , E_{m} )$ or spin variables $( \sigma_{n} , \sigma_{m} )$, and call $\Psi ( E_{n} ) \equiv C_{n}$ and $\langle E_{n} | H | E_{m} \rangle \equiv H_{n m}$. In this case, the Schrodinger equation becomes a matrix equation $$i \frac{\partial}{\partial t} C_{n} = \sum_{m} H_{n m} \ C_{m} . \ \ \ \ \ (5)$$
And finally, I leave you to work out the detail, take $( \beta , \alpha ) = ( x , p )$ and deduce the Fourier transformation between momentum and position space wave functions $$\Psi ( x ) = \int d^{3} p \ \langle x | p \rangle \ \Psi ( p ) ,$$ with $\langle x | p \rangle \sim \exp ( i p \cdot x )$ serves as transformation matrix.

Sam