$ \newcommand{\braket}[1]{\langle #1 \rangle} \newcommand{\abs}[2][]{\left\lvert#2\right\rvert_{\text{#1}}} \newcommand{\ket}[1]{\left\lvert#1 \right.\rangle} \newcommand{\bra}[1]{\langle\left. #1\right\rvert} \newcommand{\braket}[1]{\langle #1 \rangle} \newcommand{\dd}{\text{d}} \newcommand{\dv}[2]{\frac{\dd #1}{\dd #2}} \newcommand{\pdv}[2]{\frac{\partial}{\partial #1}} $
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overview
significance
derivation
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Significance

Coming soon

Derivation

Let $V_0=\infty$. Let $V(x)$ be potential energy, and $E$ be the energy of the particle. Since the particle is in the box, that means it's in a lower energy state than
We first want the possible states and associated eigenvalues for the energy operator $E$.
The operator is $H=[-\frac{\hbar}{2m}\frac{d^2}{dx^2}+V(x)]$, so we subsitute that into the eigenvector equation.
$$\begin{align*} [-\frac{\hbar}{2m}\frac{d^2}{dx^2}+V(x)]\psi(x)=E\psi(x)&=E\psi(x)\\ \implies -\frac{\hbar^2}{2m}\frac{d^2}{dx^2}\psi&=[E-V(x)]\psi(x)\\ \implies \frac{d^2}{dx^2}\psi(x)&=-\frac{2m}{\hbar^2}[E-V(x)]\psi(x). \end{align*}$$
Despite being a differential equation, this is not the same as schrodinger's equation. Solving this equation gives us the possible position states of the wavefunction and the associated energies (Schrodinger's would give the position state at a time $t$)
For the zones where $V_0=\infty$, the only solution to $$\begin{align*} \frac{d^2}{dx^2}\psi(x)=-\frac{2m}{\hbar^2}[E-\infty]\psi(x), \end{align*}$$ where $E$ is finite is $\psi(x)=0$.
For the well from $x=0$ to $x=L$, we have $V(x)=0$. That gives us $$\begin{align*} \frac{d^2}{dx}\psi(x)=-\frac{2mE}{\hbar^2}\psi(x). \end{align*}$$ If we let $k=\sqrt{\frac{2mE}{\hbar^2}}$, then we can get $$\begin{align*} \frac{d^2}{dx}\psi(x)=-k^2\psi(x). \end{align*}$$ By inspection (or solving), we can see that $\psi(x)$ is a function of $\sin(x)$ or $\cos(x)$. Let's use $\sin(x)$. Then $$\begin{align*} \psi(x)=A\sin{(kx+c)}, \end{align*}$$ where $A$ is the normalization constant, $c$ is a phase shift, and $k=\sqrt{\frac{2mE}{\hbar^2}}$ (reminder we are finding the eigenvector right now).
Right now there are two things to take care of:
  • Normalization constant $A$
  • We need the allowed energy levels such that $\psi(L)=0$ and $\psi(0)=0$ at the edges of the well, so we don't get an infinitely fast jump from $\psi(0+\epsilon)=a$ to $\psi(0-\epsilon)=0$.

We find the normalization constant. We require that $$\begin{align*} \langle E_n|E_n\rangle&=1 \\ \implies \int_{-\infty}^{\infty}\psi_n^*(x)\psi_n(x)&=1\\ &=\int_{\infty}^0\psi_n^*(x)\psi_n(x)dx+\int_{0}^L + \int_{L}^{\infty}\psi^*_n(x)\psi_n(x)dx, \end{align*}$$ Since $\psi(x)=0$ outside of $[0, L]$, then $$\begin{align*} \int_{-\infty}^{0}\psi_n^*(x)\psi_n(x)dx&=\int_{L}^{\infty}\psi_n^*(x)\psi(x)dx \\ &=0, \end{align*}$$ then we need $$\begin{align*} \int_{0}^{L}\psi_n^*\psi_n(x)dx &=1\\ \implies \int_{0}^{L}A^2\sin^2(kx)&=1\\ \implies A^2\int_{0}^{L}\sin^2(kx)&=1 \end{align*}$$ The integral of $\sin^2(kx)$ is $\frac{2x-\sin x}{4}$. Anyway if we solve for $A$, assuming that $\sin(kL)=0$ as required by the boundry coundition, then $$\begin{align*} A^2=\sqrt{\frac{2}{L}} \end{align*}$$
For the eigenstates, we need that $\psi(0)=0$ and $\psi(L)=0$. $\psi(0)=\sqrt{\frac{2}{L}}\sin(0k)=\sqrt{\frac{2}{L}}\sin(0)$, and so $c=0$. For $\psi(L)=\sqrt{\frac{2}{L}}\sin(kL)$, this will be zero when $kL$ is a multiple of $\pi$, or $k=\frac{n/pi}{L}$, where $n$ is an integer (if $n=0$, then there is no particle. If $n<0$, there is a global phase change). Hence we require $$\begin{align*} \sqrt{\frac{2mE}{\hbar^2}}=\frac{n\pi}{L} \end{align*}$$ solving for $E$ gives $$\begin{align*} E_n=\frac{\hbar^2 n^2 \pi^2}{2mL}. \end{align*}$$ Since $\hbar=\frac{h}{2\pi}$, then this can be simplified to $$\begin{align*} E_n=\frac{hn^2}{8mL^2}. \end{align*}$$ These are the allowed energy states, the energy eigenvalues.
Hence the eigenstate equation is $$\begin{align*} \frac{hn^2}{8mL^2}\sqrt{\frac{2}{L}}\psi(\frac{n\pi}{L}x)=E\psi_n(x), \end{align*}$$ where the allowed energy levels are $E_n=\frac{hn^2}{8mL^2}$ with corresponding eigenstates $\psi(x)=\sqrt{\frac{2}{L}}\sin(\frac{n\pi x}{L})$

Associated Concepts

Particle in a finite potential well

Significance

Coming soon

Derivation

Let $V_0=\infty$. Let $V(x)$ be potential energy, and $E$ be the energy of the particle. Since the particle is in the box, that means it's in a lower energy state than
We first want the possible states and associated eigenvalues for the energy operator $E$.
The operator is $H=[-\frac{\hbar}{2m}\frac{d^2}{dx^2}+V(x)]$, so we subsitute that into the eigenvector equation.
$$\begin{align*} [-\frac{\hbar}{2m}\frac{d^2}{dx^2}+V(x)]\psi(x)=E\psi(x)&=E\psi(x)\\ \implies -\frac{\hbar^2}{2m}\frac{d^2}{dx^2}\psi&=[E-V(x)]\psi(x)\\ \implies \frac{d^2}{dx^2}\psi(x)&=-\frac{2m}{\hbar^2}[E-V(x)]\psi(x). \end{align*}$$
Despite being a differential equation, this is not the same as schrodinger's equation. Solving this equation gives us the possible position states of the wavefunction and the associated energies (Schrodinger's would give the position state at a time $t$)
For the zones where $V_0=\infty$, the only solution to $$\begin{align*} \frac{d^2}{dx^2}\psi(x)=-\frac{2m}{\hbar^2}[E-\infty]\psi(x), \end{align*}$$ where $E$ is finite is $\psi(x)=0$.
For the well from $x=0$ to $x=L$, we have $V(x)=0$. That gives us $$\begin{align*} \frac{d^2}{dx}\psi(x)=-\frac{2mE}{\hbar^2}\psi(x). \end{align*}$$ If we let $k=\sqrt{\frac{2mE}{\hbar^2}}$, then we can get $$\begin{align*} \frac{d^2}{dx}\psi(x)=-k^2\psi(x). \end{align*}$$ By inspection (or solving), we can see that $\psi(x)$ is a function of $\sin(x)$ or $\cos(x)$. Let's use $\sin(x)$. Then $$\begin{align*} \psi(x)=A\sin{(kx+c)}, \end{align*}$$ where $A$ is the normalization constant, $c$ is a phase shift, and $k=\sqrt{\frac{2mE}{\hbar^2}}$ (reminder we are finding the eigenvector right now).
Right now there are two things to take care of:
  • Normalization constant $A$
  • We need the allowed energy levels such that $\psi(L)=0$ and $\psi(0)=0$ at the edges of the well, so we don't get an infinitely fast jump from $\psi(0+\epsilon)=a$ to $\psi(0-\epsilon)=0$.

We find the normalization constant. We require that $$\begin{align*} \langle E_n|E_n\rangle&=1 \\ \implies \int_{-\infty}^{\infty}\psi_n^*(x)\psi_n(x)&=1\\ &=\int_{\infty}^0\psi_n^*(x)\psi_n(x)dx+\int_{0}^L + \int_{L}^{\infty}\psi^*_n(x)\psi_n(x)dx, \end{align*}$$ Since $\psi(x)=0$ outside of $[0, L]$, then $$\begin{align*} \int_{-\infty}^{0}\psi_n^*(x)\psi_n(x)dx&=\int_{L}^{\infty}\psi_n^*(x)\psi(x)dx \\ &=0, \end{align*}$$ then we need $$\begin{align*} \int_{0}^{L}\psi_n^*\psi_n(x)dx &=1\\ \implies \int_{0}^{L}A^2\sin^2(kx)&=1\\ \implies A^2\int_{0}^{L}\sin^2(kx)&=1 \end{align*}$$ The integral of $\sin^2(kx)$ is $\frac{2x-\sin x}{4}$. Anyway if we solve for $A$, assuming that $\sin(kL)=0$ as required by the boundry coundition, then $$\begin{align*} A^2=\sqrt{\frac{2}{L}} \end{align*}$$
For the eigenstates, we need that $\psi(0)=0$ and $\psi(L)=0$. $\psi(0)=\sqrt{\frac{2}{L}}\sin(0k)=\sqrt{\frac{2}{L}}\sin(0)$, and so $c=0$. For $\psi(L)=\sqrt{\frac{2}{L}}\sin(kL)$, this will be zero when $kL$ is a multiple of $\pi$, or $k=\frac{n/pi}{L}$, where $n$ is an integer (if $n=0$, then there is no particle. If $n<0$, there is a global phase change). Hence we require $$\begin{align*} \sqrt{\frac{2mE}{\hbar^2}}=\frac{n\pi}{L} \end{align*}$$ solving for $E$ gives $$\begin{align*} E_n=\frac{\hbar^2 n^2 \pi^2}{2mL}. \end{align*}$$ Since $\hbar=\frac{h}{2\pi}$, then this can be simplified to $$\begin{align*} E_n=\frac{hn^2}{8mL^2}. \end{align*}$$ These are the allowed energy states, the energy eigenvalues.
Hence the eigenstate equation is $$\begin{align*} \frac{hn^2}{8mL^2}\sqrt{\frac{2}{L}}\psi(\frac{n\pi}{L}x)=E\psi_n(x), \end{align*}$$ where the allowed energy levels are $E_n=\frac{hn^2}{8mL^2}$ with corresponding eigenstates $\psi(x)=\sqrt{\frac{2}{L}}\sin(\frac{n\pi x}{L})$

Associated Concepts

Particle in a finite potential well
FullPage
overview
significance
derivation
related experiments
associated concepts
FullPage
overview
significance
derivation
related experiments
associated concepts

Significance

Coming soon

Derivation

Let $V_0=\infty$. Let $V(x)$ be potential energy, and $E$ be the energy of the particle. Since the particle is in the box, that means it's in a lower energy state than
We first want the possible states and associated eigenvalues for the energy operator $E$.
The operator is $H=[-\frac{\hbar}{2m}\frac{d^2}{dx^2}+V(x)]$, so we subsitute that into the eigenvector equation.
$$\begin{align*} [-\frac{\hbar}{2m}\frac{d^2}{dx^2}+V(x)]\psi(x)=E\psi(x)&=E\psi(x)\\ \implies -\frac{\hbar^2}{2m}\frac{d^2}{dx^2}\psi&=[E-V(x)]\psi(x)\\ \implies \frac{d^2}{dx^2}\psi(x)&=-\frac{2m}{\hbar^2}[E-V(x)]\psi(x). \end{align*}$$
Despite being a differential equation, this is not the same as schrodinger's equation. Solving this equation gives us the possible position states of the wavefunction and the associated energies (Schrodinger's would give the position state at a time $t$)
For the zones where $V_0=\infty$, the only solution to $$\begin{align*} \frac{d^2}{dx^2}\psi(x)=-\frac{2m}{\hbar^2}[E-\infty]\psi(x), \end{align*}$$ where $E$ is finite is $\psi(x)=0$.
For the well from $x=0$ to $x=L$, we have $V(x)=0$. That gives us $$\begin{align*} \frac{d^2}{dx}\psi(x)=-\frac{2mE}{\hbar^2}\psi(x). \end{align*}$$ If we let $k=\sqrt{\frac{2mE}{\hbar^2}}$, then we can get $$\begin{align*} \frac{d^2}{dx}\psi(x)=-k^2\psi(x). \end{align*}$$ By inspection (or solving), we can see that $\psi(x)$ is a function of $\sin(x)$ or $\cos(x)$. Let's use $\sin(x)$. Then $$\begin{align*} \psi(x)=A\sin{(kx+c)}, \end{align*}$$ where $A$ is the normalization constant, $c$ is a phase shift, and $k=\sqrt{\frac{2mE}{\hbar^2}}$ (reminder we are finding the eigenvector right now).
Right now there are two things to take care of:
  • Normalization constant $A$
  • We need the allowed energy levels such that $\psi(L)=0$ and $\psi(0)=0$ at the edges of the well, so we don't get an infinitely fast jump from $\psi(0+\epsilon)=a$ to $\psi(0-\epsilon)=0$.

We find the normalization constant. We require that $$\begin{align*} \langle E_n|E_n\rangle&=1 \\ \implies \int_{-\infty}^{\infty}\psi_n^*(x)\psi_n(x)&=1\\ &=\int_{\infty}^0\psi_n^*(x)\psi_n(x)dx+\int_{0}^L + \int_{L}^{\infty}\psi^*_n(x)\psi_n(x)dx, \end{align*}$$ Since $\psi(x)=0$ outside of $[0, L]$, then $$\begin{align*} \int_{-\infty}^{0}\psi_n^*(x)\psi_n(x)dx&=\int_{L}^{\infty}\psi_n^*(x)\psi(x)dx \\ &=0, \end{align*}$$ then we need $$\begin{align*} \int_{0}^{L}\psi_n^*\psi_n(x)dx &=1\\ \implies \int_{0}^{L}A^2\sin^2(kx)&=1\\ \implies A^2\int_{0}^{L}\sin^2(kx)&=1 \end{align*}$$ The integral of $\sin^2(kx)$ is $\frac{2x-\sin x}{4}$. Anyway if we solve for $A$, assuming that $\sin(kL)=0$ as required by the boundry coundition, then $$\begin{align*} A^2=\sqrt{\frac{2}{L}} \end{align*}$$
For the eigenstates, we need that $\psi(0)=0$ and $\psi(L)=0$. $\psi(0)=\sqrt{\frac{2}{L}}\sin(0k)=\sqrt{\frac{2}{L}}\sin(0)$, and so $c=0$. For $\psi(L)=\sqrt{\frac{2}{L}}\sin(kL)$, this will be zero when $kL$ is a multiple of $\pi$, or $k=\frac{n/pi}{L}$, where $n$ is an integer (if $n=0$, then there is no particle. If $n<0$, there is a global phase change). Hence we require $$\begin{align*} \sqrt{\frac{2mE}{\hbar^2}}=\frac{n\pi}{L} \end{align*}$$ solving for $E$ gives $$\begin{align*} E_n=\frac{\hbar^2 n^2 \pi^2}{2mL}. \end{align*}$$ Since $\hbar=\frac{h}{2\pi}$, then this can be simplified to $$\begin{align*} E_n=\frac{hn^2}{8mL^2}. \end{align*}$$ These are the allowed energy states, the energy eigenvalues.
Hence the eigenstate equation is $$\begin{align*} \frac{hn^2}{8mL^2}\sqrt{\frac{2}{L}}\psi(\frac{n\pi}{L}x)=E\psi_n(x), \end{align*}$$ where the allowed energy levels are $E_n=\frac{hn^2}{8mL^2}$ with corresponding eigenstates $\psi(x)=\sqrt{\frac{2}{L}}\sin(\frac{n\pi x}{L})$

Associated Concepts

Particle in a finite potential well

Significance

Coming soon

Derivation

Let $V_0=\infty$. Let $V(x)$ be potential energy, and $E$ be the energy of the particle. Since the particle is in the box, that means it's in a lower energy state than
We first want the possible states and associated eigenvalues for the energy operator $E$.
The operator is $H=[-\frac{\hbar}{2m}\frac{d^2}{dx^2}+V(x)]$, so we subsitute that into the eigenvector equation.
$$\begin{align*} [-\frac{\hbar}{2m}\frac{d^2}{dx^2}+V(x)]\psi(x)=E\psi(x)&=E\psi(x)\\ \implies -\frac{\hbar^2}{2m}\frac{d^2}{dx^2}\psi&=[E-V(x)]\psi(x)\\ \implies \frac{d^2}{dx^2}\psi(x)&=-\frac{2m}{\hbar^2}[E-V(x)]\psi(x). \end{align*}$$
Despite being a differential equation, this is not the same as schrodinger's equation. Solving this equation gives us the possible position states of the wavefunction and the associated energies (Schrodinger's would give the position state at a time $t$)
For the zones where $V_0=\infty$, the only solution to $$\begin{align*} \frac{d^2}{dx^2}\psi(x)=-\frac{2m}{\hbar^2}[E-\infty]\psi(x), \end{align*}$$ where $E$ is finite is $\psi(x)=0$.
For the well from $x=0$ to $x=L$, we have $V(x)=0$. That gives us $$\begin{align*} \frac{d^2}{dx}\psi(x)=-\frac{2mE}{\hbar^2}\psi(x). \end{align*}$$ If we let $k=\sqrt{\frac{2mE}{\hbar^2}}$, then we can get $$\begin{align*} \frac{d^2}{dx}\psi(x)=-k^2\psi(x). \end{align*}$$ By inspection (or solving), we can see that $\psi(x)$ is a function of $\sin(x)$ or $\cos(x)$. Let's use $\sin(x)$. Then $$\begin{align*} \psi(x)=A\sin{(kx+c)}, \end{align*}$$ where $A$ is the normalization constant, $c$ is a phase shift, and $k=\sqrt{\frac{2mE}{\hbar^2}}$ (reminder we are finding the eigenvector right now).
Right now there are two things to take care of:
  • Normalization constant $A$
  • We need the allowed energy levels such that $\psi(L)=0$ and $\psi(0)=0$ at the edges of the well, so we don't get an infinitely fast jump from $\psi(0+\epsilon)=a$ to $\psi(0-\epsilon)=0$.

We find the normalization constant. We require that $$\begin{align*} \langle E_n|E_n\rangle&=1 \\ \implies \int_{-\infty}^{\infty}\psi_n^*(x)\psi_n(x)&=1\\ &=\int_{\infty}^0\psi_n^*(x)\psi_n(x)dx+\int_{0}^L + \int_{L}^{\infty}\psi^*_n(x)\psi_n(x)dx, \end{align*}$$ Since $\psi(x)=0$ outside of $[0, L]$, then $$\begin{align*} \int_{-\infty}^{0}\psi_n^*(x)\psi_n(x)dx&=\int_{L}^{\infty}\psi_n^*(x)\psi(x)dx \\ &=0, \end{align*}$$ then we need $$\begin{align*} \int_{0}^{L}\psi_n^*\psi_n(x)dx &=1\\ \implies \int_{0}^{L}A^2\sin^2(kx)&=1\\ \implies A^2\int_{0}^{L}\sin^2(kx)&=1 \end{align*}$$ The integral of $\sin^2(kx)$ is $\frac{2x-\sin x}{4}$. Anyway if we solve for $A$, assuming that $\sin(kL)=0$ as required by the boundry coundition, then $$\begin{align*} A^2=\sqrt{\frac{2}{L}} \end{align*}$$
For the eigenstates, we need that $\psi(0)=0$ and $\psi(L)=0$. $\psi(0)=\sqrt{\frac{2}{L}}\sin(0k)=\sqrt{\frac{2}{L}}\sin(0)$, and so $c=0$. For $\psi(L)=\sqrt{\frac{2}{L}}\sin(kL)$, this will be zero when $kL$ is a multiple of $\pi$, or $k=\frac{n/pi}{L}$, where $n$ is an integer (if $n=0$, then there is no particle. If $n<0$, there is a global phase change). Hence we require $$\begin{align*} \sqrt{\frac{2mE}{\hbar^2}}=\frac{n\pi}{L} \end{align*}$$ solving for $E$ gives $$\begin{align*} E_n=\frac{\hbar^2 n^2 \pi^2}{2mL}. \end{align*}$$ Since $\hbar=\frac{h}{2\pi}$, then this can be simplified to $$\begin{align*} E_n=\frac{hn^2}{8mL^2}. \end{align*}$$ These are the allowed energy states, the energy eigenvalues.
Hence the eigenstate equation is $$\begin{align*} \frac{hn^2}{8mL^2}\sqrt{\frac{2}{L}}\psi(\frac{n\pi}{L}x)=E\psi_n(x), \end{align*}$$ where the allowed energy levels are $E_n=\frac{hn^2}{8mL^2}$ with corresponding eigenstates $\psi(x)=\sqrt{\frac{2}{L}}\sin(\frac{n\pi x}{L})$

Associated Concepts

Particle in a finite potential well
FullPage
overview
significance
derivation
related experiments
associated concepts