QM Linear Harmonic Oscillator

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We already discussed the harmonic oscillator in another article. In this article we focus on the development of an algebraic method to solve the harmonic oscillator equations. For a quadradic potential the Schrödinger Equation leads to Eigenfunctions containing the Hermite polynomials. Since, solving the second order Schrödinger Equation is rather tedious, we are interested in an easier way of solving the equations.

The Hamiltonian for the harmonic oscillator is \begin{align} H=\frac{p^2}{2m}+ \frac{1}{2}m \omega^2 x^2 \end{align} In position space we can represent the Hamiltonian as \begin{align} H= \left(\frac{ \hbar }{i }\frac{ \ d}{ dx} \right)^2 + \frac{1}{2}m \omega^2 x^2 \tag{1} \end{align} Our task is to solve the corresponding stationary Schrödinger Equation, which in position space reads \begin{align} \left( -\frac{\hbar^2}{2m}\frac{d^2 \psi}{dx^2} + \frac{1}{2}kx^2 \right) \psi(x)= E\psi(x) \end{align}


Contents

Ladder Operators

P.A.M. Dirac had the idea to factor the operator (1) in two parts \begin{align} u^2 + v^2 = (u+iv)(u-iv) \end{align} but this is not so easy since $u$ and $v$ do not commute $uv \not = vu$. First of all it's convenient to work with dimensionless operators, so let's have a look at the dimensions of the constants involved. $\hbar = [E \cdot T]$, $\omega= [T^{-1}]$, $m=[M]$ so the quantity \begin{align} \sqrt{\frac{\hbar}{m \omega}}=[L] \end{align} has units of length and \begin{align} \sqrt{\hbar m \omega}=[M \cdot L \cdot T^{-1}] \end{align} has units of momentum. This suggests to define the dimensionless operators \begin{align} a= \frac{x}{\sqrt{2\hbar/(m \omega)}} + \frac{ip}{\sqrt{2\hbar m \omega}} \end{align} and \begin{align} a^{\dagger}= \frac{x}{\sqrt{2\hbar/(m \omega)}} - \frac{ip}{\sqrt{2\hbar m \omega}} \end{align} the factor $\sqrt{2}$ we need because of the $1/2$ in the Hamilton operator. Then the product $aa^{\dagger}$ is \begin{align} aa^{\dagger}&= \left( \frac{x}{\sqrt{2\hbar/(m \omega)}} + \frac{ip}{\sqrt{2\hbar m \omega}} \right) \left( \frac{x}{\sqrt{2\hbar/(m \omega)}} - \frac{ip}{\sqrt{2\hbar m \omega}}\right)=\left( \frac{x^2}{2\hbar/(m \omega)} + \frac{p^2}{2\hbar m \omega} + \frac{x}{\sqrt{2\hbar/(m \omega)}} \frac{-ip}{\sqrt{2\hbar m \omega}} +\frac{ip}{\sqrt{2\hbar m \omega}}\frac{x}{\sqrt{2\hbar/(m \omega)}} \right)\\ &=\left( \frac{x^2}{2\hbar/(m \omega)} + \frac{p^2}{2\hbar m \omega} - \frac{[x,ip]}{2 \hbar} \right)=\left( \frac{x^2}{2\hbar/(m \omega)} + \frac{p^2}{2\hbar m \omega} + \frac{1}{2 } \mathbb{I} \right) \end{align} since $[x,ip]=[x,i]p+i[x,p]=0+ii\hbar \mathbb{I}=-\hbar \mathbb{I}$ by the rules of Operator Algebra. In the position basis this is \begin{align} aa^{\dagger}&=\left( \frac{m \omega x^2}{2\hbar} + \frac{-\hbar^2 }{2\hbar m \omega} \frac{d^2}{dx^2} + \frac{1}{2 } \mathbb{I}\right) \end{align} similarly \begin{align} a^{\dagger}a&= \left( \frac{x}{\sqrt{2\hbar/(m \omega)}} - \frac{ip}{\sqrt{2\hbar m \omega}} \right) \left( \frac{x}{\sqrt{2\hbar/(m \omega)}} + \frac{ip}{\sqrt{2\hbar m \omega}} \right) \\ &=\left( \frac{x^2}{2\hbar/(m \omega)} + \frac{p^2}{2\hbar m \omega} + \frac{[x,ip]}{2 \hbar} \right)=\left( \frac{x^2}{2\hbar/(m \omega)} + \frac{p^2}{2\hbar m \omega} - \frac{1}{2 } \mathbb{I} \right) \end{align} Notice, that the commutator of the annihilation $a$ and creation operator $a^{\dagger}$ gives exactly the unit operator \begin{align} [a,a^{\dagger}]=aa^{\dagger}-a^{\dagger}a=\mathbb{I} \end{align}

This factorisation of $H$ into two terms is a special example of an application of the general Born-Infeld Method which belongs to the research area of Supersymmetric Quantum Mechanics.



We find, that the Hamilton operator for the harmonic oscillator can be wirtten in the algebraic form \begin{align} H=\frac{p^2}{2m}+ \frac{1}{2}m \omega^2 x^2 = \hbar \omega \left( aa^{\dagger}+\frac{1}{2 } \mathbb{I}\right) \end{align} with the lowering or annihilation operator \begin{align} a= \frac{x}{\sqrt{2\hbar/(m \omega)}} + \frac{ip}{\sqrt{2\hbar m \omega}} \end{align} and the raising or creation operator \begin{align} a^{\dagger}= \frac{x}{\sqrt{2\hbar/(m \omega)}} - \frac{ip}{\sqrt{2\hbar m \omega}} \end{align} $a$ and $a^{\dagger}$ together are refered to as ladder operators.


Number Operator

Since we often work with the product $a^{\dagger}a$ we give it a name and call it the Number operator \begin{align} N=a^{\dagger}a \end{align} The Number operator is hermitian $N=N^{\dagger}$ since $(a^{\dagger}a)^{\dagger}=a^{\dagger}a^{\dagger \dagger}=aa^{\dagger}$. The commutator of $N$ with $a$ is by the rules of Operator Algebra \begin{align} [N,a]=[a^{\dagger}a,a]=a^{\dagger}[a,a]+[a^{\dagger},a]a=0+(-1)a=-a \end{align} and with $a^{\dagger}$ \begin{align} [N,a^{\dagger}]=[a^{\dagger}a,a^{\dagger}]=a^{\dagger}[a,a^{\dagger}]+[a^{\dagger},a^{\dagger}]a=a^{\dagger}(1)+0=a^{\dagger} \end{align} What are the eigenstates of the Number operator $N$? Let's assume that $|\lambda \rangle$ is an eigenvector and $\lambda$ an eigenvalue of $N$ \begin{align} N|\lambda \rangle= \lambda|\lambda \rangle \tag{2} \end{align} Now, since $N$ is hermition $\lambda$ must be real. Furthermore, we know, that $[N,a]=-a$ so \begin{align} [N,a]|\lambda \rangle= -a|\lambda \rangle\\ Na |\lambda \rangle - aN |\lambda \rangle= - a |\lambda \rangle \end{align} Using the eigenvalue equation (2) this becomes \begin{align} Na |\lambda \rangle - a \lambda |\lambda \rangle= - a |\lambda \rangle \end{align} rearranging terms we find \begin{align} Na |\lambda \rangle = a \lambda |\lambda \rangle - a |\lambda \rangle = (\lambda -1) a |\lambda \rangle \end{align} This equation tells us that

if $|\lambda \rangle$ is an eigenstate of $N$ with eigenvalue $\lambda$, then also $(a |\lambda \rangle)$ is an eigenstate of $N$ with the eigenvalue $(\lambda -1)$ lowered by one. \begin{align} N (a |\lambda \rangle) = (\lambda -1) (a |\lambda \rangle) \end{align}

We now see, why $a$ is called lowering operator, it lowers the eigenstate by one.


In exactly the same way we can apply $[N,a^{\dagger}]$ on $|\lambda \rangle$ and get \begin{align} [N,a^{\dagger}]|\lambda \rangle&= a^{\dagger}|\lambda \rangle\\ Na^{\dagger} |\lambda \rangle - a^{\dagger}N |\lambda \rangle&= a^{\dagger} |\lambda \rangle\\ Na^{\dagger} |\lambda \rangle&=(\lambda +1) a^{\dagger} |\lambda \rangle \end{align} and find that

if $|\lambda \rangle$ is an eigenstate of $N$ with eigenvalue $\lambda$, then also $(a^{\dagger} |\lambda \rangle)$ is an eigenstate of $N$ with the eigenvalue $(\lambda +1)$ raised by one. \begin{align} N (a |\lambda \rangle) = (\lambda +1) (a^{\dagger} |\lambda \rangle) \end{align}

This is why $a^{\dagger}$ is called the raising operator.

Raising and Lowering

The eigenvalues of $N$ can never be negative, which is easily proofed \begin{align} \langle \psi | N | \psi \rangle = \langle \psi | a^{\dagger} a | \psi \rangle = \| a | \psi \rangle \| \geq 0 \end{align}

This implies that there is a "lowest" state $| 0 \rangle$, such that \begin{align} a | 0 \rangle = 0 | 0 \rangle \end{align} $a$ does not lower this state, otherwise the state would become negative. This state is called ground state. In Field Theory, $| 0 \rangle$ is called vacuum.

Let us use the $|n \rangle$ as notation for the eigenstates of $N$. Applying $N$ on $|n \rangle$ just gives a real number $n$, that's also why $N$ is called the Number operator. \begin{align} N|n \rangle= n|n \rangle \end{align} Now we can write the Schrödinger Equation as \begin{align} i\hbar \frac{d}{dt} | n \rangle = H | n \rangle = \hbar \omega (N + \frac{1}{2} \mathbb{I}) | n \rangle \end{align} and \begin{align} \hbar \omega (N + \frac{1}{2} \mathbb{I}) | n \rangle =\hbar \omega (n + \frac{1}{2}) | n \rangle \end{align} In the ground state $n=0$ and we have \begin{align} H | n \rangle=\hbar \omega \frac{1}{2} | n \rangle \end{align} the Energy in the ground state is not zero, but $E=\frac{1}{2}\hbar \omega$. Now, starting from the ground state we successively apply $a^{\dagger}$ and raise the states to $| 1 \rangle,| 2 \rangle, \dots | n \rangle$. Applying $a^{\dagger}$ on $| 0 \rangle$ we have \begin{align} |1 \rangle=c \cdot a^{\dagger} | 0 \rangle \end{align} $|1 \rangle$, but to normalise $|1 \rangle$ we also need to multiply by a normalisation constant $c$. In general we have \begin{align} |n+1 \rangle=c_n \cdot a^{\dagger} | n \rangle \end{align} and we need to find out what $c_n$ is. Normalising $|n+1 \rangle$ we have \begin{align} \langle n+1 |n+1 \rangle=1\\ |c_n|^2 \langle n| a^{\dagger} a| n \rangle=1 \end{align} From the commutation relation we know that $aa^{\dagger}-a^{\dagger}a=1$, so $a^{\dagger}a=N+1$ thus \begin{align} |c_n|^2 \langle n|N+1| n \rangle=|c_n|^2 \langle n|n+1| n \rangle=|c_n|^2 (n+1) \langle n| n \rangle=1 \end{align} and we finally have for the coefficient \begin{align} c_n=\frac{1}{\sqrt{n+1}} \end{align} with this factor we can raise and preserve normalisation with \begin{align} |n+1 \rangle=\frac{1}{\sqrt{n+1}} a^{\dagger} | n \rangle \end{align} a similar calculation for the lowering operator gives \begin{align} |n-1 \rangle=\frac{1}{\sqrt{n}} a | n \rangle \end{align} Applying the raising operator $n$ times on the ground state we get \begin{align} |n \rangle=\frac{(a^{\dagger})^n}{\sqrt{n!}} | 0 \rangle \end{align}

Let us summerize these important results

\begin{align} \sqrt{n+1}|n+1 \rangle=a^{\dagger} | n \rangle \end{align} \begin{align} \sqrt{n}|n-1 \rangle= a | n \rangle \end{align} \begin{align} |n \rangle=\frac{(a^{\dagger})^n}{\sqrt{n!}} | 0 \rangle \end{align}

Position Space Wave Functions

Let us apply the lowering operator on the ground state \begin{align} a|0 \rangle= 0 \end{align} writing $a$ in terms of $x$ and $p$ the same equation reads \begin{align} \left( \frac{x}{\sqrt{2\hbar/(m \omega)}} + \frac{ip}{\sqrt{2\hbar m \omega}} \right) |0 \rangle= 0 \end{align} but we can get rid of some factors by multiplying through by $\sqrt{2\hbar m \omega}$. To represent this equation in position space we multiply on the left by $\langle x|$

\begin{align} \langle x|\left( m \omega x + ip \right) |0 \rangle= 0 \end{align} $x$ operating on $\langle x|$ just gives a scalar $x$, which we can pull in front. $p$ in position space is $-i\hbar d/dx$, which can be pulled in front as well, leaving

\begin{align} \left( m \omega x - \hbar \frac{d}{dx} \right) \langle x|0 \rangle= 0 \end{align} $\langle x|0 \rangle$ is just the ground state wave equation in position space $\phi_0 (x)$. This way we have found a first oder differential equation for the ground state wave function, that is much easier to solve, than the second order Hermite differential equation. The solution of

\begin{align} \frac{d}{dx} \phi_0 (x) = \frac{ m \omega x}{\hbar } \phi_0 (x) \end{align} is gaußian wave in $x$

\begin{align} \phi_0 (x) =C e^{-\int dx' \frac{ m \omega x'}{\hbar }}=C e^{\frac{- m \omega x^2}{2\hbar }} \end{align} Normalising $\phi_0 (x)$ by solving the Gaußian Integral we find the constant $C$ \begin{align} \int_{-\infty}^{\infty} dx|\phi_0 (x)|^2 =\int_{-\infty}^{\infty} dx C^2 e^{\frac{ -m \omega x^2}{\hbar }} =C^2 \sqrt{\frac{\pi \hbar}{m \omega}} = 1 \end{align} Therefore the coefficient is \begin{align} C=\left(\frac{m \omega}{\pi \hbar} \right)^{1/4} \end{align}


The Ground State Position Space Wave Function is

\begin{align} \phi_0 (x) =\left(\frac{m \omega}{\pi \hbar} \right)^{1/4} e^{\frac{ -m \omega x^2}{2\hbar }} \end{align}


Now we are ready to apply the raising operator in the position space \begin{align} \langle x|\frac{1}{\sqrt{1}} \left( \frac{x}{\sqrt{2\hbar/(m \omega)}} - \frac{ip}{\sqrt{2\hbar m \omega}} \right) |0 \rangle = \left( \sqrt{\frac{m \omega}{2\hbar}} x - \frac{\hbar}{\sqrt{2\hbar m \omega}} \frac{d}{dx} \right)\phi_0 (x) \end{align} so we get $\phi_1 (x) $ by simply differentiating a gaußian; this is of course again a gaußian with an additional factor. However we also multiply this function by $x$. Since the gaußian is a symmetric function $\phi_1 (x) $ will thus become an antisymmetric function. Applying $a^{\dagger}$ $n$-times we have \begin{align} \phi_n (x)=\frac{1}{\sqrt{n!}} (a^{\dagger})^n \phi_0 (x)=\frac{1}{\sqrt{ n!}} \left( \frac{x}{\sqrt{2\hbar/(m \omega)}} - \frac{ip}{\sqrt{2\hbar m \omega}} \right)^n \left(\frac{m \omega}{\pi \hbar} \right)^{1/4} e^{\frac{- m \omega x^2}{2\hbar }} \end{align}

After varibale transformation $\zeta = m \omega x^2$ the raising operator can be written

\begin{align} \left( \frac{x}{\sqrt{2\hbar/(m \omega)}} - \frac{ip}{\sqrt{2\hbar m \omega}} \right):=\frac{1}{\sqrt{2}} \left( \zeta - \frac{d}{d \zeta} \right) \end{align} The general solution $\phi_n (x) $ is of the form \begin{align} \phi_n (x)=\frac{1}{\sqrt{n!}} (a^{\dagger})^n \phi_0 (x)=\frac{1}{\sqrt{ n!}} \left( \zeta - \frac{d}{d \zeta} \right)^n \left(\frac{m \omega}{\pi \hbar} \right)^{1/4} e^{\frac{-\zeta^2}{2\hbar }} = H_n (\zeta) e^{\frac{-\zeta^2}{2\hbar }} \end{align} where $H_n$ is the Hermite polynomial of n-th order. It also holds in general that $\phi_n (x)$ is an even function for $n$ even and odd if $n$ is odd. We can determine $H_n$ directly by the Rodriges Formula \begin{align} H_n(\zeta)= (-1)^n e^{\zeta^2} \frac{d^n}{dx^n} e^{-\zeta^2} \end{align} They satisfy the orthonormality relation

\begin{align} \int_{-\infty}^{\infty}d\zeta \, H_m(\zeta)H_n(\zeta)= \sqrt{\pi} \, 2^n n! \delta_{mn} \end{align}


Uncertainties

We can also find relations for $x$ and $p$ in terms of the ladder operators. We set $m, \omega, \hbar$ equal to one.

\begin{align} x = \frac{a + a^{\dagger}}{\sqrt{2}} \end{align} \begin{align} p = \frac{a - a^{\dagger}}{\sqrt{2}} \end{align}

To find the uncertainty e.g. in $x$ \begin{align} \Delta x = \sqrt{ \langle x^2 \rangle - \langle x \rangle^2 } \end{align}

we need to find $\langle x^2 \rangle$ and $\langle x \rangle$ \begin{align} \langle x \rangle = \langle n | x | n \rangle = 0\\ \langle p \rangle = \langle n | p | n \rangle = 0\\ \end{align} The mean value of the position and the momentum is zero.

\begin{align} \langle x^2 \rangle &= \langle n | x^2 | n \rangle\\ &= \langle n | \left( \frac{a + a^{\dagger}}{2} \right)^2 | n \rangle\\ &= \langle n | \left( \frac{a^2 + (a^{\dagger})^2 +aa^{\dagger}+a^{\dagger}a}{2} \right) | n \rangle\\ \end{align} Raising twice gives $\sqrt{(n+1)(n+2)}\langle n | n+2 \rangle$, but $\langle n | n+2 \rangle$ are orthogonal. The same argument applies to the lowering opperater squared, so we can cancel the squared operators \begin{align} &= \langle n | \left( \frac{aa^{\dagger}+a^{\dagger}a}{2} \right) | n \rangle\\ \end{align} $aa^{\dagger}+a^{\dagger}a=N+1+N$ \begin{align} &= \langle n | \left(\frac{ N+1+N}{2} \right)| n \rangle=n+\frac{1}{2}\\ \end{align} restoring units by adding units of lenght squared, we have \begin{align} \langle x^2 \rangle &= (n+\frac{1}{2}) \frac{\hbar}{m \omega} \end{align} Similarly we find for $p$ \begin{align} \langle p^2 \rangle &= (n+\frac{1}{2}) \hbar m \omega \end{align} so we find for

the uncertainty of position and momentum in eigenstate $|n \rangle$ \begin{align} (\Delta x_n)(\Delta p_n) =(n+\frac{1}{2}) \hbar \end{align}

The ground state is the state of minimum uncertainty \begin{align} (\Delta x_0)(\Delta p_0) =\frac{\hbar}{2} \end{align}


Coherent States

What are the Eigenstates of $a$ and $a^{\dagger}$? To save time we again get rid of constants $m, \omega, \hbar$ by setting them to one. Then the ladder operators read \begin{align} a^{\dagger}=\frac{x-ip}{\sqrt{2}} \end{align} and \begin{align} a=\frac{x+ip}{\sqrt{2}} \end{align} In the position basis they become \begin{align} a^{\dagger}=\frac{1}{\sqrt{2}} \left( x - i \frac{d}{dx} \right) \end{align} and \begin{align} a=\frac{x+ip}{\sqrt{2}}=\frac{1}{\sqrt{2}} \left( x + i \frac{d}{dx} \right) \end{align}

If we apply $a^{\dagger}$ on the eigenstate, which we call $\chi(x)$ with eigenvalue $\lambda$ we have

\begin{align} \frac{1}{\sqrt{2}} \left( x - i \frac{d}{dx} \right)\chi(x)= \lambda \chi(x) \end{align} but the eigenstate $\chi(x)=e^{x-\lambda}C$ is not normalisable.

The ladder operator $a^{\dagger}$ does not have normalisable eigenstates.

However applying $a$ to the eigenstate $\langle x|\alpha \rangle=\alpha(x)$ in the position basis, with eigenvalue $\alpha$ \begin{align} \frac{1}{\sqrt{2}} \left( x + i \frac{d}{dx} \right)\alpha(x)= \alpha \alpha(x) \end{align} we do find normalisable eigenstates. Now, let us expand $|\alpha \rangle$ in the basis $\{|n \rangle \}$ \begin{align} |\alpha \rangle = \sum_{n=0}^{\infty} c_n |n \rangle =c_0|0 \rangle +c_1|1 \rangle +c_2|2 \rangle +c_3|3 \rangle + \dots \end{align} If we apply $a$ on alpha we get \begin{align} a|\alpha \rangle = \sum_{n=0}^{\infty} c_n \sqrt{n+1}|n+1 \rangle =c_0\sqrt{1}|1 \rangle +c_1\sqrt{2}|2 \rangle +c_2\sqrt{3}|3 \rangle +c_3\sqrt{4}|4 \rangle + \dots \end{align} \begin{align} a|\alpha \rangle = \sum_{n=0}^{\infty} c_n \alpha |n \rangle =\alpha(c_0|0 \rangle +c_1|1 \rangle +c_2|2 \rangle +c_3|3 \rangle + \dots) \end{align} Matching terms on the RHS we find \begin{align} c_1&= \frac{\alpha}{\sqrt{1}}c_0\\ c_2&= \frac{\alpha}{\sqrt{2}}c_1\\ c_3&= \frac{\alpha}{\sqrt{3}}c_2\\ & \vdots \end{align} and in general \begin{align} c_n= \frac{\alpha^n}{\sqrt{n!}}c_0\\ \end{align} Now we can express $|\alpha \rangle$ as \begin{align} | \alpha \rangle = \sum_{n=0}^{\infty} c_n |n \rangle = c_0 \sum_{n=0}^{\infty} \frac{\alpha^n}{\sqrt{n!}} |n \rangle \tag{3} \end{align} and determine $c_0$ by the normalisation condition. \begin{align} \langle \alpha |\alpha \rangle = |c_0|^2 \sum_{n=0}^{\infty} \frac{|\alpha|^{2n}}{n!} \langle n|n \rangle =1 \end{align} By definition the infinite sum is the exponential $e^{|\alpha|^2}$, so we find \begin{align} c_0 = e^{-\frac{1}{2}|\alpha|^2} \end{align} Using (3) together with the relation for $|n \rangle$ in terms of the ground state \begin{align} |n \rangle = \frac{(a^{\dagger})^n}{\sqrt{n!}} |0 \rangle \end{align} we can write $| \alpha \rangle $ as \begin{align} | \alpha \rangle = e^{-\frac{1}{2}|\alpha|^2} \sum_{n=0}^{\infty} \frac{(\alpha a^{\dagger})^n}{n!} |0 \rangle \end{align} again using the definition of $e^{\alpha a^{\dagger}}$ as an infinite series we get \begin{align} | \alpha \rangle = e^{-\frac{1}{2}|\alpha|^2} e^{\alpha a^{\dagger}} |0 \rangle \end{align} we call $| \alpha \rangle$ a Coherent state. Let us investigate the commutator \begin{align} [\alpha a^{\dagger}, \alpha ^* a]=\alpha a^{\dagger}\alpha ^* a -\alpha ^* a\alpha a^{\dagger}=\alpha ^*\alpha (1+N-N)= \alpha ^* \alpha= |\alpha|^2 \end{align} This gives us a relation for $|\alpha|^2$. Using this and the rules of Operator Algebra, we can write $|\alpha \rangle$ as \begin{align} | \alpha \rangle = e^{-\frac{1}{2}[\alpha a^{\dagger}, \alpha ^* a]} e^{\alpha a^{\dagger}} |0 \rangle = e^{\frac{1}{2}(\alpha a^{\dagger}-\alpha ^* a +[[\alpha a^{\dagger}, \alpha ^* a],\alpha a^{\dagger}]/2} |0 \rangle= e^{\alpha a^{\dagger}-\alpha ^* a} |0 \rangle \end{align}

Let us summarise our findings.

A Coherent State or Quasi-Classical State$ | \alpha \rangle$ is an eigenstate of the annihilation operator $a$ with eigenvalue $\alpha$. A Coherent state can be expanded in the basis $\{ |n \rangle \}$, where it can be written in the neat and compact form. \begin{align} | \alpha \rangle = e^{\alpha a^{\dagger}-\alpha ^* a} |0 \rangle \end{align} $a$ is not hermitian, so $\alpha \in \mathbb{C}$. In specific, the ground state with $\alpha=0$ is a coherent state.






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