Wave function using the time dependent Schrodinger equation

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SUMMARY

The discussion focuses on deriving the wave function ψ(x, t) of a free particle using the time-dependent Schrödinger equation. The equation is defined as $$\mathrm{i} \hbar \partial_t \psi(x,t)=\hat{H} \psi(x,t)$$ with the Hamiltonian $$\hat{H}=\frac{\hat{p}^2}{2m} + V(\hat{x})$$. The formal solution is expressed as $$\psi(x,t)=\exp \left (-\frac{\mathrm{i} \hat{H} t}{\hbar} \right) \psi(x,0)$$, which can be rewritten using the propagator $$G(x,x',t)$$. For a free particle, the propagator is evaluated using momentum eigenstates, leading to the expression $$G(x,x',t)=\int_{\mathbb{R}} \mathrm{d} p \langle x |\exp[-\mathrm{i} \hat{p}^2 t/(2m \hbar)]|p \rangle \langle p|x' \rangle$$.

PREREQUISITES
  • Understanding of the time-dependent Schrödinger equation
  • Familiarity with Fourier transforms in quantum mechanics
  • Knowledge of Hamiltonian mechanics, specifically $$\hat{H}=\frac{\hat{p}^2}{2m} + V(\hat{x})$$
  • Concept of propagators and Green's functions in quantum theory
NEXT STEPS
  • Study the derivation and properties of Green's functions in quantum mechanics
  • Learn about the application of Fourier transforms to solve the Schrödinger equation
  • Explore the concept of momentum eigenstates and their role in quantum mechanics
  • Investigate the implications of time evolution in quantum systems using the propagator
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Students and professionals in quantum mechanics, physicists working with wave functions, and researchers interested in the time evolution of quantum systems.

ThiagoSantos
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Given a wavefunction ψ(x, 0) of a free particle at initial time t=0, I need to write the general expression of the function at time t. I used a Fourier transform of ψ(x, t) in terms of ψ(p, t), but, i don't understand how to use green's functions and the time dependent Schrödinger equation to get my answer. What's the relationtship between them?
 
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The time-dependent Schrödinger equation reads
$$\mathrm{i} \hbar \partial_t \psi(x,t)=\hat{H} \psi(x,t).$$
Let's assume for simplicity that
$$\hat{H}=\frac{\hat{p}^2}{2m} + V(\hat{x}),$$
i.e., that ##\hat{H}## is not explicitly time dependent. Then the formal solution of the equation above is
$$\psi(x,t)=\exp \left (-\frac{\mathrm{i} \hat{H} t}{\hbar} \right) \psi(x,0).$$
This you can write in the form
$$\psi(x,t)=\int_{\mathbb{R}} \mathrm{d} x' \left \langle x \left |\exp \left (-\frac{\mathrm{i} \hat{H} t}{\hbar} \right) \right| x' \right \rangle \psi(x',0)= \int_{\mathbb{R}} \mathrm{d} x' G(x,x',t) \psi(x',0),$$
i.e., the propator is
$$G(x,x',t)=\left \langle x \left |\exp \left (-\frac{\mathrm{i} \hat{H} t}{\hbar} \right) \right| x' \right \rangle.$$
Usually it's of course difficult to really calculate the propagator.

For a free particle, where ##\hat{H}=\hat{p}^2/(2m)## you can use the momentum eigenstates to evaluate it:
$$G(x,x',t)=\int_{\mathbb{R}} \mathrm{d} p \langle x |\exp[-\mathrm{i} \hat{p}^2 t/(2m \hbar)]|p \rangle \langle p|x' \rangle.$$
 

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