Phy5645/One dimensional problem: Difference between revisions

Revision as of 16:06, 7 August 2013

Let us start with the original box, with its walls at $x=0\!$ and $x=a,\!$ and with the particle in the ground state of this box. The energy and the wave function are

$E_{0}={\frac {\pi ^{2}\hbar ^{2}}{2ma^{2}}}$

and

$\psi _{0}(x)={\sqrt {\frac {2}{a}}}\sin \left({\frac {\pi x}{a}}\right),\,0<x<a.$

(a) In the new box, with the right-hand wall now located at $x=4a,\!$ the ground state energy and wave function of the electron are

$E'_{0}={\frac {\pi ^{2}\hbar ^{2}}{32ma^{2}}}$

and

$\psi '_{0}(x)={\frac {1}{\sqrt {2a}}}\sin \left({\frac {\pi x}{4a}}\right).$

The probability of finding the electron in $\psi _{0}(x)\!$ is

$P(E'_{0})=\left|\int _{0}^{a}\psi _{0}^{\ast }(x)\psi '_{0}(x)\,dx\right|^{2}={\frac {1}{a^{2}}}\left|\int _{0}^{a}\sin \left({\frac {\pi x}{a}}\right)\sin \left({\frac {\pi x}{4a}}\right)\,dx\right|^{2}.$

Note that the upper limit of the integral is $a;\!$ this is because $\psi _{0}(x)\!$ is limited to the region between 0 and $a.\!$ Using the identity,

$\sin {a}\sin {b}={\tfrac {1}{2}}[\cos(a+b)-\cos(a-b)],$

we get

$P(E'_{0})={\frac {1}{4a^{2}}}\left|\int _{0}^{a}\cos \left({\frac {5\pi x}{4a}}\right)\,dx-\int _{0}^{a}\cos \left({\frac {3\pi x}{4a}}\right)\,dx\right|^{2}$

$={\frac {128}{225\pi ^{2}}}=0.058=5.8\%.$

(b) The energy and wave function of the first excited state of the new box are

$E'_{1}={\frac {\pi ^{2}\hbar ^{2}}{8ma^{2}}}$

and

$\psi '_{1}(x)={\frac {1}{\sqrt {2a}}}\sin \left({\frac {\pi x}{2a}}\right).$

The probability of finding the particle in this state is

$P(E'_{2})=\left|\int _{0}^{a}\psi _{0}^{\ast }(x)\psi '_{1}(x)\,dx\right|^{2}={\frac {1}{a^{2}}}\left|\int _{0}^{a}\sin \left({\frac {\pi x}{a}}\right)\sin({\frac {\pi x}{2a}})\right|^{2}dx$

$={\frac {16}{9\pi ^{2}}}=0.18=18\%.$

Back to Summary of One-Dimensional Systems

@@ Line 1: / Line 1: @@
-Let us start with the old box, with its walls at <math>x=0\!</math> and <math>x=a,\!</math> and with the particle in the ground state of this box.  The energy and wavefunction are
+Let us start with the original box, with its walls at <math>x=0\!</math> and <math>x=a,\!</math> and with the particle in the ground state of this box.  The energy and the wave function are
-<math>E_{0}=-\frac{\pi ^{2}\hbar^{2}}{2ma^{2}}</math>
+<math>E_{0}=\frac{\pi ^{2}\hbar^{2}}{2ma^{2}}</math>
 and
@@ Line 7: / Line 7: @@
 <math>\psi_0(x)=\sqrt{\frac{2}{a}}\sin\left (\frac{\pi x}{a}\right ),\, 0<x<a.</math>
-'''(a)'''  Now in the new box i.e., x=a and x=4a, the ground state energy and wave function of the electron are<math>
+'''(a)'''  In the new box, with the right-hand wall now located at <math>x=4a,\!</math> the ground state energy and wave function of the electron are
-{E_{1}}'=-\frac{\pi ^{2}\hbar^{2}}{2m\left ( 4a \right )^{2}}=  -\frac{\pi ^{2}\hbar^{2}}{32ma^{2}}</math>
+<math>E'_{0}=\frac{\pi ^{2}\hbar^{2}}{32ma^2}</math>
 and
-<math>\psi _{1}\left ( x \right )= \frac{1}{\sqrt{2a}} sin\left ( \frac{\pi x}{4a} \right )</math>
+<math>\psi'_{0}(x)= \frac{1}{\sqrt{2a}}\sin\left (\frac{\pi x}{4a}\right ).</math>
-The probability of finding the electron in <math>\psi _{1}\left ( x \right )</math> is
+The probability of finding the electron in <math>\psi_{0}(x)\!</math> is
-<math>P\left ( {E}'_{1} \right )=\left | \langle\psi _{1}|\phi _{1} \right \rangle^{2}= \left | \int_{0}^{a}\psi _{1}\ast \left ( x \right )\phi _{1}(x) \right |^{2}dx= \frac{1}{a^{2}}\left | \int_{0}^{a}sin\left ( \frac{\pi x}{4a}\right ) sin(\frac{\pi x}{a})\right |^{2}dx</math>
+<math>P(E'_{0})=\left |\int_{0}^{a}\psi_{0}^{\ast}(x)\psi'_{0}(x)\,dx\right |^2=\frac{1}{a^{2}}\left |\int_{0}^{a}\sin\left (\frac{\pi x}{a}\right )\sin\left (\frac{\pi x}{4a}\right )\,dx\right |^2.</math>
-the upper limit of the integral sign is 'a' because <math>\phi _{1}(x)</math> is limited to the region between 0 and a. Using the relation <math>sin\left ( a \right )sin(b)=\frac{1}{2}cos(a+b)-\frac{1}{2}cos(a-b)</math>, we get
+Note that the upper limit of the integral is <math>a;\!</math> this is because <math>\psi_{0}(x)\!</math> is limited to the region between 0 and <math>a.\!</math>  Using the identity,
-<math>P\left ( {E}'_{1} \right )=\frac{1}{a^{2}}\left | \frac{1}{2}\int_{0}^{a}cos\left ( \frac{5\pi x}{4a} \right )dx-\frac{1}{2}\int_{0}^{a}cos\left ( \frac{3\pi x}{4a} \right )dx \right |^{2}dx</math>
+<math>\sin{a}\sin{b}=\tfrac{1}{2}[\cos(a+b)-\cos(a-b)],</math>
-<math>=\frac{128}{\left ( 15 \right )^{2}\pi ^{2}}=0.058=5.8 percent</math>
+we get
-'''(b)''' If the electron is in the first excited state of the new box, its energy and wavefunctions are
+<math>P(E'_{0})=\frac{1}{4a^{2}}\left |\int_{0}^{a}\cos\left (\frac{5\pi x}{4a}\right )\,dx-\int_{0}^{a}\cos\left ( \frac{3\pi x}{4a}\right )\,dx\right |^2</math>
-<math>P\left ( {E}'_{2} \right )=\left | \langle\psi _{2}|\phi _{1} \right \rangle^{2}= \left | \int_{0}^{a}\psi _{2}\ast \left ( x \right )\phi _{1}(x) \right |^{2}dx= \frac{1}{a^{2}}\left | \int_{0}^{a}sin\left ( \frac{\pi x}{2a}\right ) sin(\frac{\pi x}{a})\right |^{2}dx</math>
+<math>=\frac{128}{225\pi ^{2}}=0.058=5.8%.</math>
-<math>=\frac{16}{9\pi ^{2}}=0.18=18percent</math>
+'''(b)''' The energy and wave function of the first excited state of the new box are
+<math>E'_{1}=\frac{\pi ^{2}\hbar^{2}}{8ma^2}</math>
+and
+<math>\psi'_{1}(x)= \frac{1}{\sqrt{2a}}\sin\left (\frac{\pi x}{2a}\right ).</math>
+The probability of finding the particle in this state is
+<math>P(E'_{2})=\left |\int_{0}^{a}\psi_{0}^{\ast}(x)\psi'_{1}(x)\,dx\right |^2=\frac{1}{a^2}\left |\int_{0}^{a}\sin\left ( \frac{\pi x}{a}\right )\sin(\frac{\pi x}{2a})\right |^{2}dx</math>
+<math>=\frac{16}{9\pi ^{2}}=0.18=18%.</math>
 Back to [[Summary of One-Dimensional Systems]]

Phy5645/One dimensional problem: Difference between revisions

Revision as of 16:06, 7 August 2013

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