particle in an infinite well Archives - Time Travel, Quantum Entanglement and Quantum Computing https://stationarystates.com/tag/particle-in-an-infinite-well/ Not only is the Universe stranger than we think, it is stranger than we can think...Hiesenberg Fri, 10 Oct 2025 15:10:57 +0000 en-US hourly 1 https://wordpress.org/?v=6.9.1 Particle in an Infinite 1D Square Well — Energy States https://stationarystates.com/basic-quantum-theory/particle-in-an-infinite-1d-square-well-energy-states/?utm_source=rss&utm_medium=rss&utm_campaign=particle-in-an-infinite-1d-square-well-energy-states https://stationarystates.com/basic-quantum-theory/particle-in-an-infinite-1d-square-well-energy-states/#comments Fri, 10 Oct 2025 15:06:55 +0000 https://stationarystates.com/?p=1021 Energy States of a Particle in an Infinite 1D Square Well Setup Consider a well of width with potential     The time-independent Schrödinger equation (TISE) inside the well is […]

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Energy States of a Particle in an Infinite 1D Square Well

Setup

Consider a well of width a with potential

    \[ V(x)= \begin{cases} 0, & 0<x<a,\\ \infty, & \text{otherwise.} \end{cases} \]

The time-independent Schrödinger equation (TISE) inside the well is

    \[ -\frac{\hbar^2}{2m}\,\frac{d^2\psi}{dx^2}=E\,\psi, \qquad 0<x<a, \]

with boundary conditions \psi(0)=\psi(a)=0 (the wavefunction vanishes where V=\infty).

Solve the TISE

Let k\equiv \sqrt{2mE}/\hbar. Then \psi''+k^2\psi=0 has the general solution

    \[ \psi(x)=A\sin(kx)+B\cos(kx). \]

Apply the boundary conditions:

  • \psi(0)=0 \Rightarrow B=0.
  • \psi(a)=0 \Rightarrow A\sin(ka)=0.

For a nontrivial solution A\neq 0, we require

    \[ \sin(ka)=0 \;\Rightarrow\; ka=n\pi,\quad n=1,2,3,\dots \]

Hence

    \[ k_n=\frac{n\pi}{a},\qquad E_n=\frac{\hbar^2 k_n^2}{2m} =\frac{n^2\pi^2\hbar^2}{2ma^2}. \]

Normalized Eigenfunctions

Normalization \int_0^a |\psi_n|^2\,dx=1 yields

    \[ \psi_n(x)=\sqrt{\frac{2}{a}}\;\sin\!\left(\frac{n\pi x}{a}\right), \qquad n=1,2,3,\dots,\quad 0<x<a. \]

Key Properties

Quantized energies (non-degenerate):

    \[ E_n=\frac{n^2\pi^2\hbar^2}{2ma^2},\quad n=1,2,3,\dots \]

Level spacing grows with n:

    \[ E_{n+1}-E_n=\frac{(2n+1)\pi^2\hbar^2}{2ma^2}. \]

Orthonormality:

    \[ \int_0^a \psi_n(x)\,\psi_m(x)\,dx=\delta_{mn}. \]

Nodes: \psi_n has n-1 interior nodes. Higher n means more oscillations.

Expectation values (stationary states):

    \[ \langle x\rangle=\frac{a}{2},\qquad \langle p\rangle=0,\qquad \langle p^2\rangle=2mE_n=\left(\frac{n\pi\hbar}{a}\right)^2. \]

Time dependence: For an energy eigenstate, the full solution is

    \[ \Psi_n(x,t)=\psi_n(x)\,e^{-iE_n t/\hbar}. \]

Alternative Well Placement (-L/2,\,L/2)

If the well runs from -L/2 to +L/2, the spectrum is the same with a\to L, but the eigenfunctions split into definite parity. A convenient labeling is

    \[ \psi_{n}^{(\text{even})}(x) =\sqrt{\frac{2}{L}}\cos\!\left(\frac{(2n-1)\pi x}{L}\right), \quad n=1,2,3,\dots \]

    \[ \psi_{n}^{(\text{odd})}(x) =\sqrt{\frac{2}{L}}\sin\!\left(\frac{2n\pi x}{L}\right), \quad n=1,2,3,\dots \]

and the energies remain

    \[ E_n=\frac{n^2\pi^2\hbar^2}{2mL^2}\quad (\text{with appropriate } n). \]



 


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