In 1-D problems there is a theorem which says,
Ground state has no node, 1\(^{\text{st}}\) excited state has one node 2\(^{\text{nd}}\) excited state has 2 nodes.
so Question is, why does the no. of nodes increase as Energy increases.
To have wave fun. normalizable. It should be clamped at both ends.
\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=1.0]
% ground state
\begin{scope}[yshift=0cm]
\draw[thick] (0,-0.25) -- (0,0.25);
\draw[thick] (3.0,-0.25) -- (3.0,0.25);
\draw[thick] (0,0) .. controls (0.9,1.05) and (2.1,1.05) .. (3.0,0);
\node[anchor=west] at (3.9,0.5) {$0$ nodes};
\node[anchor=west] at (6.0,0.5) {Ground state};
\end{scope}
% first excited
\begin{scope}[yshift=-1.9cm]
\draw (0,0) -- (3.0,0);
\draw[thick] (0,-0.25) -- (0,0.25);
\draw[thick] (3.0,-0.25) -- (3.0,0.25);
\draw[thick] (0,0) .. controls (0.5,1.0) and (1.0,1.0) .. (1.5,0)
.. controls (2.0,-1.0) and (2.5,-1.0) .. (3.0,0);
\node[below] at (1.5,-0.5) {node};
\node[anchor=west] at (3.9,0.3) {$\underline{1}$ node};
\node[anchor=west] at (6.0,0.3) {1$^{\text{st}}$ excited state};
\end{scope}
% second excited
\begin{scope}[yshift=-3.9cm]
\draw (0,0) -- (3.0,0);
\draw[thick] (0,-0.25) -- (0,0.25);
\draw[thick] (3.0,-0.25) -- (3.0,0.25);
\draw[thick] (0,0) .. controls (0.35,0.85) and (0.65,0.85) .. (1.0,0)
.. controls (1.35,-0.85) and (1.65,-0.85) .. (2.0,0)
.. controls (2.35,0.85) and (2.65,0.85) .. (3.0,0);
\node[anchor=west] at (3.9,0.3) {$2$ node};
\node[anchor=west] at (6.0,0.3) {3$^{\text{rd}}$ excited state};
\end{scope}
\end{tikzpicture}
It is because of curvature keeps changing as energy increases. \(\phi''(x)\) contributes more and more as curvature changes.
(curvature (1-D))
So we can see that how \(\nabla^2\) contributes to increase of Energy eigen value.
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\begin{tikzpicture}[scale=1.0]
\draw[->] (-3.4,0) -- (5.0,0) node[right] {$x$};
\draw[->] (-1.1,-1.6) -- (-1.1,2.9) node[above] {$V(x)$};
\draw[thick] (-3.3,0.05) .. controls (-2.9,0.1) and (-2.75,1.9) .. (-2.4,1.9)
.. controls (-2.05,1.9) and (-1.9,-0.2) .. (-1.5,-1.1)
.. controls (-1.2,-1.55) and (-0.7,-1.55) .. (-0.35,-1.05)
.. controls (-0.05,-0.6) and (0.05,0.85) .. (0.35,0.85)
.. controls (0.6,0.85) and (0.7,0.2) .. (0.95,0.15)
.. controls (1.35,0.15) and (1.6,2.6) .. (2.0,2.6)
.. controls (2.5,2.6) and (2.9,0.35) .. (3.6,0.12)
.. controls (4.2,0.05) and (4.6,0.03) .. (4.9,0.03);
\draw (-3.3,1.55) -- (4.4,1.55) node[right] {$E_4$};
\draw (-3.3,0.35) -- (4.6,0.35) node[right] {$E_3$};
\draw[dashed] (-3.0,-0.72) -- (0.1,-0.72) node[right] {$E_2$};
\draw[dashed] (-2.6,-1.45) -- (-0.7,-1.45);
\node[right] at (-0.6,-1.45) {$E_1$};
\end{tikzpicture}
let \(E_i\) be total energy of system. \(\{E_i\} = \{E_1, E_2, E_3, \cdots\}\)
\((E_1 < 0)\) and \(E_1 < V(\min)\) :--
Such case does not exist as we will see that \(\langle\)K.E\(\rangle\) is always a real positive no.
Since \(x, p\) does not commute, \([x,p] \neq 0\) we can not have definite value of K.E and P.E simultaneously in the same state. as.
(or)
In general the eigen states of \(\hat H\) are not the eigen states of K.E or the P.E separately.
(or)
In a stationary state, the particle can not have definite value either of its K.E or its P.E. But only sum of the two has definite value of measurement.
So we can't say given the total energy so much is K.E and so much is P.E.
In arbitrary state \(|\Psi\rangle\).
\(\langle\)K.E\(\rangle\) or \(\langle\hat p^2\rangle\) is given by
(as \(\hat p\) is hermitian \((\hat p = \hat p^{\dagger})\), \((\hat p^2 = \hat p^{\dagger}\hat p)\))
(length of vector)
so \(\langle\)K.E\(\rangle\) is always a positive real no. i.e. \(\langle\)K.E\(\rangle \geq 0\).
so we can't have total energy lesser than P.E because that would imply \(\langle\)K.E\(\rangle < 0\).
"A particle can not have definite total energy when it is at perticular place. because \([\hat H, \hat x] \neq 0\).
also,
\((E_2)\) \(\big(E_2 > V(x)\big)\) :
The wave function looks like this :--
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\begin{tikzpicture}[scale=1.0]
\draw (-2.6,0) -- (2.8,0);
\draw[thick] (-1.0,1.3) .. controls (-0.75,0.1) and (-0.4,-1.15) .. (0.0,-1.2)
.. controls (0.4,-1.15) and (0.75,0.1) .. (1.0,1.3);
\draw[thick] (-2.4,0.08) .. controls (-1.5,0.2) and (-0.9,1.1) .. (-0.35,1.15)
.. controls (0.2,1.15) and (0.9,0.25) .. (2.6,0.06);
\node[below right] at (0.15,-0.05) {$E_2$};
\end{tikzpicture}
Most of the time we will find particle inside the well, but there is a finite non-zero probability that it can exist outside the well (tunneling).
\((E_3)\) & \((E_4)\) :--
can we have such energy value / level?
Not strictly, These can not be strict stationary states or bound states. Because of the finite width of potential barrier \((\Delta V)\), there exists finite probability of the particle to exist outside the barrier / (tunnel through barrier).
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\begin{tikzpicture}[scale=1.0]
\draw[->] (-3.4,0) -- (4.4,0) node[right] {$x$};
\draw[->] (-1.2,-1.4) -- (-1.2,3.0) node[above] {$V(x)$};
\draw[thick] (-3.2,0.1) .. controls (-2.7,0.2) and (-2.6,2.2) .. (-2.2,2.25)
.. controls (-1.8,2.2) and (-1.6,-0.5) .. (-1.2,-1.2)
.. controls (-0.85,-1.5) and (-0.4,-1.1) .. (-0.15,-0.5)
.. controls (0.05,0.0) and (0.1,0.5) .. (0.35,0.55)
.. controls (0.75,0.55) and (1.2,2.6) .. (1.7,2.6)
.. controls (2.3,2.6) and (2.8,0.3) .. (3.6,0.1)
.. controls (4.0,0.06) and (4.2,0.05) .. (4.3,0.05);
\draw (-2.9,1.0) -- (3.4,1.0) node[right] {$E_4$};
\draw[dashed] (0.35,0.55) -- (0.35,1.0);
\draw[<->] (0.35,-0.35) -- (2.6,-0.35);
\node[below] at (1.5,-0.4) {$\frac{\Delta V}{\Delta x}$};
\draw[<->] (1.7,1.05) -- (1.7,2.55);
\node[right] at (1.75,1.8) {$H$};
\draw (0.75,1.0) circle (0.42);
\node[anchor=west,align=left] at (2.4,-1.1) {potential barrier of finite width\\ and finite height.};
\end{tikzpicture}
so \(\exists \int|\Psi(x)|^2 > 0\) for \(x >\) barrier width.
so there can't be fixed / bound state.
\(E_4\)) If we place particle in such potential. initially. As time goes along this state would evolve under such \(\hat H\), such that we end up having finite probability to find it outside the well.
Note :--
If we place particle near origin well. and iff we have very large width and height; then we have tunneling probability to be exceedingly low. And we might find the particle inside this well for very -- very long time but eventually it is bound to escape. Such a state is called metastable state.
"Probability of tunneling becomes zero iff the barrier has finite width and infinite height"
Suppose we have delta fun \((\delta(x))\) barrier, could particle tunnel through it?
Yes because width is infinitensimally small and height is infinite such that their product is finite.
If we have width of barrier infinite from either side, then we can't have \(E_2\) energy level. But still it should have finite probability to tunnel to nearby finite barrier wells. (?)(?)(?).
How it is reflected in the eigen states / eigen functions of \(\hat H\).
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\begin{tikzpicture}[scale=1.0]
\draw[->] (-2.6,0) -- (2.6,0);
\draw[->] (0,-1.4) -- (0,2.4) node[above] {$V(x)$};
\draw[thick] (-1.7,2.0) .. controls (-0.9,0.1) and (-0.4,0.0) .. (0,0)
.. controls (0.4,0.0) and (0.9,0.1) .. (1.7,2.0);
\node[anchor=west] at (0.15,-0.75) {$\times$ (E)};
\node[anchor=west,align=left] at (0.35,-1.25) {$\uparrow$\\ (not allowed)};
\end{tikzpicture}
\(V(x) \neq V(-x)\) say,
So what kind of eigen states or energy eigen value you would expect?
\((E < 0\) not allowed\()\)
\(E > 0\) allowed. for (stationary / eigen states of \(\hat H\)) and discrete set of energy eigen values.
For we such potential
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\begin{tikzpicture}[scale=1.0]
\draw[->] (-3.0,0) -- (3.2,0) node[right] {$x$};
\draw[->] (0,-1.5) -- (0,2.2) node[above] {$V(x)$};
\draw[thick] (-2.8,1.3) .. controls (-1.4,1.3) and (-0.9,-1.15) .. (0,-1.2)
.. controls (0.9,-1.15) and (1.4,1.3) .. (2.8,1.3);
\draw (-2.6,1.3) -- (2.6,1.3);
\node[above left] at (-0.05,1.3) {$V_0$};
\foreach \y in {-0.9,-0.55,-0.2,0.15} {
\draw[gray] (-1.15,\y) -- (1.15,\y);
}
\node[anchor=west,align=left] at (3.4,1.1) {we will have (for $E < V_0$)\\ discrete set of eigen\\ values (in the well)\\ and contineous set of\\ eigen values outside the\\ well. $(E > V_0)$};
\end{tikzpicture}
For \(E > V_0\) eigen functions are non -- normalizable.
Each eigen function must obey :--
let \(x' = -x\), or \((x = -x')\)
eq\(^{\text{n}}\) becomes.
that would the proved \(V(-x') = V(x')\) (or odd)
"since there is no degeneracy in 1-D, so every eigen value should have unique eigen fun / state.
\(\{|\Psi\rangle\) & \(\alpha|\Psi\rangle\) are same thing in Q.M
or
so,
If potential has symmetry (reflection / mirror symmetry) / Parity invariance then solutions have definite parity. (either odd / even parity).
If potential is even function, wave function must be even or odd, but it can't be a mixed function.
What would the ground state be? (even or odd)?
Even because, if it were odd, wave fun must be contineous and because it is an odd fun it must have zero value at \(x = 0\), so it means it has node. Ground state does not have node.
so ground state will be even function.
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\begin{tikzpicture}[scale=1.0]
\draw (-3.4,0) -- (3.4,0) node[right] {$x$};
\draw[->] (0,-0.4) -- (0,3.4) node[above] {$V(x)$};
\draw[thick] (-1.75,3.1) parabola bend (0,0.1) (1.75,3.1);
\draw (-2.6,0.75) -- (1.55,0.75) node[right] {$E_0$ Ground state};
\draw (-2.6,1.7) -- (1.4,1.7) node[right] {$E_1$};
\draw (-2.6,2.6) -- (1.25,2.6) node[right] {$E_2$};
% ground state wavefn
\draw[thick] (-1.3,0.75) .. controls (-0.7,0.78) and (-0.45,1.55) .. (0,1.55)
.. controls (0.45,1.55) and (0.7,0.78) .. (1.3,0.75);
% first excited
\draw[thick] (-1.5,1.7) .. controls (-1.15,1.75) and (-0.9,2.45) .. (-0.55,2.45)
.. controls (-0.2,2.45) and (0.0,1.7) .. (0.0,1.7)
.. controls (0.0,1.7) and (0.2,0.95) .. (0.55,0.95)
.. controls (0.9,0.95) and (1.15,1.65) .. (1.5,1.7);
\node[anchor=west] at (3.6,3.0) {(2 nodes) even function};
\node[anchor=west] at (3.6,1.9) {(1$^{\text{st}}$ excited) odd functions};
\node[anchor=west] at (3.6,0.9) {even function};
\node[anchor=west] at (3.9,3.5) {\underline{Eigen function}};
\end{tikzpicture}
\[
[\hat H, \hat P] = 0
\]
\[
\Rightarrow \quad \text{All eigen functions of } \hat H \text{ are eigen functions of parity operator.}
\]
But converse is not true (All odd/even functions are not eigen fun. of this perticular \(\hat H\)). (simultaneously.
eigen fun. of \(\hat H\) are subset of eigen fun. of Parity operator.
Question :-- what kind of eigen functions / eigen values we expect?
let the general
\(\hookrightarrow\) for it to be (\(V(x)\) even fun.)
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\begin{tikzpicture}[scale=1.0]
\draw[->] (-2.6,0) -- (2.8,0) node[right] {$(x)$};
\draw[->] (0,-0.3) -- (0,2.4) node[above] {$V(x)$};
\draw[thick,domain=-1.45:1.45,samples=80] plot ({\x},{\x*\x});
\draw[thick,domain=-2.1:2.1,samples=80] plot ({\x},{abs(\x)});
\draw[thick,domain=-2.4:2.4,samples=120] plot ({\x},{sqrt(abs(\x))*1.35});
\node[anchor=west] at (1.1,2.15) {$r = 2$};
\node[anchor=west] at (1.55,1.6) {$r = 1$};
\node[anchor=west] at (2.0,1.95) {$r = \tfrac{1}{2}$};
\node[anchor=west,align=left] at (3.4,1.2) {In all of such pote-\\ -ntial we expect\\ bound state with\\ discrete energy levels.\\ and wave functions to\\ be eigen odd / even functions};
\end{tikzpicture}
\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=1.0]
\draw[->] (-2.9,0) -- (2.9,0) node[right] {$x$};
\draw[->] (0,-2.0) -- (0,2.2) node[above] {$p$};
\draw[thick] (0,1.6) ellipse [x radius=1.9, y radius=0] ;
\draw[thick] (-1.9,0) .. controls (-1.85,1.1) and (-0.9,1.6) .. (0,1.6)
.. controls (0.9,1.6) and (1.85,1.1) .. (1.9,0)
.. controls (1.85,-1.1) and (0.9,-1.6) .. (0,-1.6)
.. controls (-0.9,-1.6) and (-1.85,-1.1) .. (-1.9,0);
\draw (-1.9,0) -- (0,1.6) -- (1.9,0) -- (0,-1.6) -- cycle;
\draw[dashed] (-1.9,1.6) -- (1.9,1.6);
\draw[dashed] (-1.9,-1.6) -- (1.9,-1.6);
\draw[dashed] (-1.9,-1.7) -- (-1.9,1.7);
\draw[dashed] (1.9,-1.7) -- (1.9,1.7);
\node[anchor=east] at (-1.95,1.75) {$E^{1/2}$};
\node[anchor=west] at (0.1,1.75) {$\sqrt{2mE}$};
\node[anchor=west] at (2.0,0.75) {$\sim E^{1/r}$};
\end{tikzpicture}
lets find out semiclassically.
lets find,
so, area goes like
for large quantum no \((n)\) we expect results of quantum theory to be matched with semiclassical result.
for \(r = 2\)
so we expect that for \(n \gg 1\) energy levels are proportional to '\(n\)' itself
for \(r = 1\)
and
as, \(r \to \infty\)
(This is what happens in 1-D well). with \(\infty\) potential barriers
what would happen if \(V(x) = \lambda(\cosh x - 1)\)
\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=1.0]
\draw[->] (-3.2,0) -- (3.4,0) node[right] {$x$};
\draw[->] (0,-0.5) -- (0,3.4) node[above] {$V(x)$};
\draw[thick,domain=-2.0:2.0,samples=120] plot ({\x},{(exp(\x)+exp(-\x))/2 - 1});
\draw[thick,domain=-2.4:1.15,samples=120] plot ({\x},{exp(\x)});
\draw[thick,domain=-1.15:2.4,samples=120] plot ({\x},{exp(-\x)});
\node[anchor=east] at (-0.35,3.15) {$(\cosh x - 1)$};
\draw[->] (-0.35,3.05) -- (-0.95,2.45);
\node[anchor=east] at (-1.15,2.05) {$e^{-x}$};
\node[anchor=west] at (1.15,3.15) {$e^{x}$};
\end{tikzpicture}
[\(\cosh x = \dfrac{e^{x}+e^{-x}}{2}\)]
So how does energy values should go like.
\(E_n\) must increase slower than \(n^2\) as \(n\uparrow\) and faster than \(\left(n^{2-\varepsilon}\right)\) for arbitary small \(\varepsilon\).
\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=1.0]
\draw[->] (-2.7,0) -- (2.7,0) node[right] {$x$};
\draw[->] (0,-1.8) -- (0,2.0);
\node[anchor=east] at (-0.1,1.85) {$p$};
\draw[thick] (0,0) ellipse [x radius=1.6, y radius=1.1];
\node[anchor=west] at (0.55,1.8) {$E^{1/2}$};
\draw[->] (0.55,1.7) -- (0.08,1.2);
\node[anchor=north] at (1.0,-1.15) {$\ln E$};
\draw[->] (1.05,-1.05) -- (1.45,-0.5);
\end{tikzpicture}
\[
\frac{p^2}{2m} + \lambda(\cosh x - 1) = E
\]
\[
e^{x_0} \gg \frac{p^2}{2m}
\]
\[
E \sim e^{x_0}
\]
\((x_0 \sim \ln E)\)
[(struck out) [?]]
but \(\ln(\ln E)\) is much smaller
so
using first root in eq (1)
(Ignoring factor of 2)
Finally lets solve
In position basis \(\hat p\) is differential operator \(\left(-i\hbar\frac{d}{dx}\right)\)
Note :-
"The energy eigen values are independent of what basis we choose.
Eigen values & corresponding eigen states are independent of the basis we choose to solve the problem."
Note :- It is not possible to write \(\hat x, \hat p\) in terms of finite-diamentional matrices. as we can see, \((xp - px = i\hbar I)\)
[\(N =\) matrix diamention]
"we need infinite diamentions to represent operators if we choose to use matrix representation."
For use of differential operator then there is no questions of matrices & traces.
For eigen value of energy \(E\), & eigen state \(|\phi(x)\rangle\)
we already know that \(E \geq 0\)
using boundary cond\(^{\text{n}}\)
after that
redefine \(|\phi(x)\rangle\) such that
since eq\(^{\text{n}}\) (1) is non-linear in '\(x\)', principle of superposition does not imply here, so we have to solve it using different method.
Eq. (1) is a kind of Hermites equation.
Paul dirac broke the non-linear \(\hat H\) into two linear operators (which commute with each other).
[1st order factors]
such that \(\hat H\,\phi(x) = E\,\phi(x)\)
(How to factorize \(x^2+p^2\) \(\Rightarrow\) \((x+ip)(x-ip)\) thats what exactly dirac did.).
[\(x+ip\) is not allowed because of diamencional reasons. so we have to make it diamensionless.]
(1)
1\(^{\text{st}}\) step is to get rid of '\(i\hbar\)' in below eq\(^{\text{n}}\).
parameters in problem are \(m, \omega, \hbar\)
so quantity of length can be written as
Note :- we have to make factors diamentionless, as '\(x+ip\)' for '\(x-ip\)' is not allowed
[\(\nearrow\) diamentionless]
lets define operator
and
such that / now
since \(a \neq a^{\dagger}\), \(a\) is not hermitian operator
so its not a physical observable.
similarly
what is
[\(\to\) unit operator]
[Heisenberg algebra: \([a,a] = 0 = [a^{\dagger},a^{\dagger}]\) ; \([a^{\dagger},a] = I\)]
"\(a^{\dagger}a\) is not finite diamentional." as
Define \(a^{\dagger}a = N\) (number operator has eigen values will turn out to be \(0,1,2\cdots\) Natural numbers)
\(\left\{\begin{aligned}&[A,BC] = [A,B]C + B[A,C]\\ &[AB,C] = [A,C]B + A[B,C]\end{aligned}\right\}\) commutator rules
so, \([N,a]\) becomes
similarly
Now our job is to find eigen-states of
since eigen states of \(\hat I\) is trivial (any fun. is eigen-state)
we have to look for eigen states of '\(a^{\dagger}a\)'.
"All eigen-states of \(a^{\dagger}a\) will be eigen states of \(\hat H\)"
Let, \(|\lambda\rangle\) be eigen state of \(N\) \((a^{\dagger}a)\) with eigen value \(\lambda\).
\(\therefore\) '\(\lambda\)' has to be real as \(a^{\dagger}a\) is hermitian.
we know, \([N,a] = -a\)
so, lets apply to same eigen state \(|\lambda\rangle\)
[\(\hat a\) is linear operator \(a\lambda|\lambda\rangle = \lambda(a|\lambda\rangle)\)]
This implies that if \(|\lambda\rangle\) is eigen state of \(\hat N\). so is the \(\hat a|\lambda\rangle\) an eigen state of \(\hat N\). but with eigen value reduced by unity.
exactly same way
if \(|\lambda\rangle\) is eigen state of \(\hat N\), so is the \(a^{\dagger}|\lambda\rangle\) an eigen state of \(\hat N\), with eigen value increased by 1.
so, \(a^{\dagger}\) is called raising operator,
and \(a\) is called as lowering operator.
suppose we have state (eigen) \(|\lambda\rangle\) of operator \(\hat N\). lets see about \(a|\lambda\rangle\) state, then \(a|\lambda\rangle\) state is also an eigen state of \(\hat N\) but with reduced eigen value by 1.
| *eigen state of \(\hat N\)* | eigen value of state | Energy |
| \(\vert\lambda\rangle\) | \(\lambda\) | \(\hbar\omega(\lambda)\) |
| \(a\vert\lambda\rangle\) | \(\lambda-1\) | \(\hbar\omega(\lambda-1)\) |
| \(a^2\vert\lambda\rangle\) | \(\lambda-2\) | \(\hbar\omega(\lambda-2)\) |
| \(\vdots\) | \(\vdots \quad \geq 0\) | \(\vdots \quad \geq 0\) |
eventually \(\big(a^{n}|\lambda\rangle\big) = \lambda - n\) can hit negative.
But \(\hat N\) can not have negative eigen values.
after all, \(\langle N\rangle =\) arithmetic avg of all eigen values.
so no eigen values should be negative.
so, \(\lambda\) has to be positive integers.
(or)
\(\exists\) a state \(|0\rangle\) such that,
so there must be a state with eigen values zero such that \(a|0\rangle = 0\). and we can't lower the eigen states & values further.
'\(|0\rangle\)' is called ground state. in Q.M
—————————— vacuum in Q.F.T.
and anhilation operator \(a\) anhiletes the vaccum.
\(|0\rangle\) is not a null vector here. it is a eigen state \(|\phi_0(x)\rangle\) whose eigen value is zero.
To Find eigen states \(|n\rangle\).
[let stationary states of \(\hat H\) is represented as \(|n\rangle\).]
so \(\hbar\omega\left(n+\frac{1}{2}\right)\) are energy levels.
Ground state
1\(^{\text{st}}\) excited state,
2\(^{\text{nd}}\) excited state,
\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=1.0]
\draw (-2.9,0) -- (4.2,0);
\draw (0,-0.5) -- (0,3.5);
\draw[thick,domain=-1.8:1.8,samples=120] plot ({\x},{\x*\x});
\draw (-2.5,0.5) -- (1.8,0.5) node[right] {$\hbar\omega/2$};
\draw (-2.5,1.5) -- (1.8,1.5) node[right] {$3\hbar\omega/2$};
\draw (-2.5,2.5) -- (1.8,2.5) node[right] {$5\hbar\omega/2$};
\draw[<->] (3.8,1.5) -- (3.8,2.5) node[midway,right] {$\hbar\omega$};
\draw[<->] (3.8,0.5) -- (3.8,1.5) node[midway,right] {$\hbar\omega$};
\draw[<->] (3.8,0.0) -- (3.8,0.5) node[midway,right] {$\hbar\omega/2$};
\node[anchor=west] at (5.3,1.5) {$\Big\}$ equispaced};
\end{tikzpicture}
lets say we have find ground state \(|0\rangle\), which is normalized. \((\langle 0|0\rangle = 1)\)
lets find 1\(^{\text{st}}\) excited state.
lets put normalization const. \(c\).
In general.
\(\Big(aa^{\dagger} - a^{\dagger}a = \mathbb{1}\)
\(aa^{\dagger} - N = I \Rightarrow (aa^{\dagger} = N+I)\)
[assuming \(\langle n|n\rangle = 1\)]
so
In exactly same way,
for \(n=0\), \(a|0\rangle = 0\) and there is no state \(\underline{|-1\rangle}\)
lets see how the wave function looks like.
representation of ground state in position basis.
so
\(\Big\}\) no nodes, gaussian
Now, \((A_0 = ?)\)
Now lets find \(|1\rangle\).
[\(|n\rangle = \frac{1}{\sqrt{n}}a^{\dagger}|n-1\rangle\)]
(We don't have to normelize it again, it is previously normalized)
since we have \(\phi_0(x)\), we insert it in above eq\(^{\text{n}}\) and get normelized \(\phi_1(x)\).
we can see that \(\phi_0(x)\) is even function, and \(\phi_1(x)\) is odd function.
\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=1.0]
\draw[->] (-2.6,0) -- (7.4,0) node[right] {$x$};
% phi_0
\draw[->] (0,-0.7) -- (0,1.8);
\node[anchor=west] at (0.15,1.75) {$\phi_0(x)$};
\draw[thick,domain=-2.2:2.2,samples=120] plot ({\x},{1.3*exp(-\x*\x)});
% phi_1
\begin{scope}[xshift=4.6cm]
\draw[->] (0,-1.7) -- (0,1.8);
\node[anchor=west] at (0.15,1.75) {$\phi_1(x)$};
\draw[thick,domain=-2.2:2.2,samples=140] plot ({\x},{-3.0*\x*exp(-\x*\x)});
\end{scope}
\end{tikzpicture}
\[
\boxed{\phi_n(x) = A_n\,H_n\!\left(x\sqrt{\frac{m\omega}{2\hbar}}\right)e^{-\frac{m\omega x^2}{2\hbar}}}
\]
\[
E_n = \hbar\omega\left(n+\frac{1}{2}\right) \qquad (n = 0,1,2\cdots\infty).
\]