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Solving Harmonic Oscillator Problems using the Operator Method: Definition of Ladder Operators 📂Quantum Mechanics

Solving Harmonic Oscillator Problems using the Operator Method: Definition of Ladder Operators

Build-up

The ladder operators of the harmonic oscillator are represented by Ladder Operator\mathrm{Ladder\ Operator}. The Hamiltonian, which is the energy operator, HH can also be substituted, and the characteristic functions of the ladder operators can be used to find the eigenfunctions from the ground state. Hints for defining new operators can be obtained by factorizing the classical Hamiltonian of the harmonic oscillator HH.

H=12mp2+12mmw2x2=12m(p2+m2w2x2)=12m(ip+mwx)(−ip+mwx) \begin{align*} H &= \frac{1}{2m}p^{2}+\frac{1}{2m}mw^{2}x^{2} \\ &= \frac{1}{2m} \left(p^{2}+m^{2}w^{2}x^{2} \right) \\ &= \frac{1}{2m} (ip+mwx)(-ip+mwx) \end{align*}

From this hint, the two ladder operators of the harmonic oscillator can be defined as follows. The reason ii is attached to pp rather than xx and why the preceding constant is 12ℏmw\dfrac{1}{\sqrt{2\hbar mw}} is to simplify the calculation considering the final form of the equation.

a+=12ℏmw(ip+mwx)a−=12ℏmw(−ip+mwx)=(a+)∗ \begin{align*} a_{+} &= \frac{1}{\sqrt{2\hbar mw}}(ip+mwx) \\ a_{-} &= \frac{1}{\sqrt{2\hbar mw}}(-ip+mwx)=(a_{+})^{\ast} \end{align*}

Now, let’s represent the Hamiltonian HH as the product of two operators. Be cautious, as the result can vary depending on the order of multiplication since they are operators. Use the standard commutation relation [x,p]=iℏ[x,p]=i\hbar.

a−a+=12ℏmw(p2+m2w2x2+mwipx−mwixp)=12ℏmw(p2+m2w2x2−mwi[x,p])=12ℏmw(p2+2w2x2+mwℏ)=12ℏmw(p2+m2w2x2)+12ℏmw(mwℏ)=1ℏw12m(p2+m2w2x2)+12=1ℏwH+12 \begin{align*} a_{-}a_{+} &= \frac{1}{2\hbar mw} (p^{2} + m^{2}w^{2}x^{2}+mwipx-mwixp) \\ &= \frac{1}{2\hbar mw} (p^{2} + m^{2}w^{2}x^{2}-mwi[x,p]) \\ &= \frac{1}{2\hbar mw} (p^{2} + ^{2}w^{2}x^{2}+mw\hbar) \\ &= \frac{1}{2\hbar mw} (p^{2} + m^{2}w^{2}x^{2}) + \frac{1}{2\hbar mw}(mw\hbar) \\ &= \frac{1}{\hbar w}\frac{1}{2m} (p^{2} + m^{2}w^{2}x^{2}) + \frac{1}{2} \\ &= \frac{1}{\hbar w}H+\frac{1}{2} \end{align*}

Therefore, it is H=ℏw(a−a+−12)H = \hbar w(a_{-}a_{+} - \dfrac{1}{2}). By reversing the order of multiplication and following the same process, we find

a+a−=1ℏwH−12  ⟹  H=ℏw(a+a−+12) \begin{align*} && a_{+}a_{-} &= \dfrac{1}{\hbar w}H-\dfrac{1}{2} \\ \implies && H &= \hbar w(a_{+}a_{-} + \frac{1}{2}) \end{align*}

Using this, the commutator of the two operators can also be calculated. With this result, we can represent the Schrödinger equation of the harmonic oscillator using the two new operators [a−,a+]=1[a_{-},a_{+}]=1.

Hψ=Eψ H\psi=E\psi

  ⟹  ℏw(a±a∓±12)ψ=Eψ \implies \hbar w (a_\pm a_\mp \pm \dfrac{1}{2})\psi=E\psi

The names of these two operators a+a_{+} and a−a_{-} are respectively raising operator rasing operator\mathrm{rasing\ operator} and lowering operator lowering operator\mathrm{lowering\ operator}. The reason for these names is that when applied to the eigenfunctions, energy (eigenvalue) increases or decreases, respectively. In other words, they are operators that raise or lower energy. For further details, refer to the next post: Solving the Harmonic Oscillator Problem Using the Operator Method: Applying Ladder Operators.

Previous post: Solving the Harmonic Oscillator Problem Using the Operator Method: Definition of Ladder Operators.

Let’s explore how ladder operators act on the eigenfunctions of the harmonic oscillator. For reference, the commutator of an arbitrary constant and operator is always 00. The relation used in the following expansion is $( [AB,C]=A[B,C]+[A,C]B,\ \ [a_{-},a_{+}]=1 )

H=(a+a−+12)ℏw H=(a_{+}a_{-}+\dfrac{1}{2})\hbar w이므로,

[H,a+]=[(a+a−+12)ℏw,a+]=[a+a−ℏw,a+]+[12ℏw,a+]=ℏw[a+a−,a+]=ℏw(a+[a−,a+]+[a+,a+]a−)=ℏwa+=Ha+−a+H \begin{align*} [H,a_{+}] &= [(a_{+}a_{-}+\dfrac{1}{2})\hbar w,a_{+}] \\ &= [a_{+}a_{-}\hbar w,a_{+}]+[\dfrac{1}{2}\hbar w, a_{+}] \\ &= \hbar w[a_{+}a_{-},a_{+}] \\ &= \hbar w(a_{+}[a_{-},a_{+}] + [a_{+},a_{+}]a_{-}) \\ &= \hbar w a_{+} \\ &= Ha_{+} - a_{+}H \end{align*}

같은 방법으로 [H,a−] [H, a_{-}]를 구하면[H,a−]=−ℏwa−=Ha−−a−H [H, a_{-}]=- \hbar w a_{-}=Ha_{-} - a_{-}H 이제 위에서 구한 관계식을 써서 조화진동자의 고유함수에 적용시켜보자.a± a_\pm 는 고유함수∣ψ⟩ \ket{\psi} 에 대해서 고유값 방정식을 만족시키는 연산자가 아니므로a± a_\pm 대신 H H 가 고유함수에 적용될 수 있도록 모양을 바꿔준다.슈뢰딩거 방정식은 $ H \ket{\psi}=E \ket{\psi}.

Ha+∣ψ⟩=(ℏwa++a+H)∣ψ⟩=(ℏwa++a+E)∣ψ⟩=(E+ℏw)a+∣ψ⟩ \begin{align*} Ha_{+} \ket{\psi} &= (\hbar wa_{+} + a_{+}H) \ket{\psi} \\ &= (\hbar wa_{+} + a_{+}E) \ket{\psi} \\ &= (E+\hbar w)a_{+} \ket{\psi} \end{align*}

Therefore, when ∣ψ⟩\ket{\psi} is an eigenfunction of HH, a+∣ψ⟩a_{+} \ket{\psi} also satisfies the eigenvalue equation. In this case, the eigenvalue of a+∣ψ⟩a_{+} \ket{\psi} is (E+ℏw)(E+\hbar w). To generalize, let’s apply it twice. Use the result of applying it once here.

H(a+)2∣ψ⟩=Ha+a+∣ψ⟩=(a+H+ℏwa+)a+∣ψ⟩=a+H+a+∣ψ⟩+ℏwa+a+∣ψ⟩=a+(E+ℏw)a+∣ψ⟩+ℏwa+a+∣ψ⟩=(E+ℏw)(a+)2∣ψ⟩+ℏw(a+)2∣ψ⟩=(E+2ℏw)(a+)2∣ψ⟩ \begin{align*} H(a_{+})^{2} \ket{\psi} &= Ha_{+}a_{+} \ket{\psi} \\ &= (a_{+}H+\hbar w a_{+})a_{+} \ket{\psi} \\ &= a_{+}H+a_{+} \ket{\psi} + \hbar w a_{+}a_{+} \ket{\psi} \\ &= a_{+}(E+\hbar w)a_{+} \ket{\psi} + \hbar w a_{+}a_{+} \ket{\psi} \\ &= (E+\hbar w)(a_{+})^{2} \ket{\psi} + \hbar w (a_{+})^{2} \ket{\psi} \\ &= (E+2\hbar w)(a_{+})^{2} \ket{\psi} \end{align*}

Thus, by applying a+a_{+} to the eigenfunction ∣ψ⟩\ket{\psi} nn times, we obtain the following result.

  ⟹  H(a+)n∣ψ⟩=(E+nℏw)(a+)n∣ψ⟩ \implies H(a_{+})^n \ket{\psi}=(E+n\hbar w)(a_{+})^n \ket{\psi}

Similarly, by applying a−a_{-} to the eigenfunction in the same manner,

Ha−∣ψ⟩=(a−H−ℏwa−)∣ψ⟩=(E−ℏw)a−∣ψ⟩ Ha_{-} \ket{\psi}=(a_{-}H - \hbar w a_{-}) \ket{\psi}=(E-\hbar w)a_{-} \ket{\psi}

  ⟹  H(a−)n∣ψ⟩=(E−nℏw)(a−)n∣ψ⟩\implies H(a_{-})^n \ket{\psi}=(E-n\hbar w)(a_{-})^n \ket{\psi}

Therefore, the operator a+a_{+} that increases energy when applied to the eigenfunction is called the raising operator, rasing operator\mathrm{rasing\ operator}. The operator a−a_{-} that decreases energy when applied is called the lowering operator, lowering operator\mathrm{lowering\ operator}. Next post: Energy Levels and Ground State of the Harmonic Oscillator.

Previous post: Applying Ladder Operators to Solve the Harmonic Oscillator Problem.

Let’s continue to find the eigenfunctions of the energy and ground state of the harmonic oscillator.

E=<ψ∣H∣ψ⟩=<ψ∣(a+a−+12)ℏw∣ψ⟩ E=<\psi|H \ket{\psi}=<\psi|(a_{+}a_{-}+\dfrac{1}{2})\hbar w \ket{\psi}

In this case, you can’t indefinitely apply a−a_{-} to the eigenfunctions. Energy cannot be less than 00. Since there is no solution when energy is less than the potential, E>UE>U it must be. In other words, there is a ground state where the energy level no longer decreases ground state\mathrm{ground\ state} and when the lowering operator lowering operator\mathrm{lowering\ operator} is applied to the ground state, 00 occurs. Therefore, if we denote the ground state as ∣ψ0⟩ \ket{\psi_{0}}, then a−∣ψ0⟩=0a_{-} \ket{\psi_{0}}=0. Now let’s use this fact to find the ground state energy.

H∣ψ0⟩=(a+a−+12)ℏw∣ψ0⟩=ℏwa+a−∣ψ0⟩+12ℏw∣ψ0⟩=12ℏw∣ψ0⟩ \begin{align*} H \ket{\psi_{0}} &= (a_{+}a_{-} + \frac{1}{2})\hbar w \ket{\psi_{0}} \\ &= \hbar w a_{+}a_{-} \ket{\psi_{0}}+\frac{1}{2}\hbar w \ket{\psi_{0}} \\ &= \frac{1}{2}\hbar w \ket{\psi_{0}} \end{align*}

\therefore H \ket{\psi_{0}}=\dfrac{1}{2}\hbar w \ket{\psi_{0}} 바닥상태의에너지는바닥상태의 에너지는 E_{0}=\dfrac{1}{2}\hbar w 이다.이다.0이아니다!이전글에서사다리연산자는고유함수의에너지를이 아니다!이전 글에서 사다리 연산자는 고유함수의 에너지를 \pm \hbar w 만큼변화시키는것을알았다.즉,첫번째들뜬상태의에너지는만큼 변화시키는 것을 알았다.즉, 첫 번째 들뜬 상태의 에너지는 E_{1}=\dfrac{1}{2}\hbar w +\hbar w 두번째들뜬상태의에너지는두 번째 들뜬 상태의 에너지는 E_2=\dfrac{1}{2}\hbar w + 2\hbar w 따라서따라서 n 번째에너지에대해서일반적으로표현할수있다.번째 에너지에 대해서 일반적으로 표현할 수 있다. E_{n}=(n+\dfrac{1}{2})\hbar w,\ \ (n=0,\ \ 1,\ \ 2,\ \ \cdot)에너지준위가등간격(에너지 준위가 등간격( \hbar w )으로이루어져있음을알수있다.이제바닥상태의고유함수)으로 이루어져있음을 알 수 있다.이제 바닥상태의 고유함수 \psi_{0} 를구체적으로구해보자.참고로운동량연산자를 구체적으로 구해보자.참고로 운동량 연산자 p $는 p=ℏi∂∂x p={\hbar \over i}{\partial \over \partial x} $ a_{-}\psi_{0}=0.

  ⟹  12ℏmw(−ip+mwx)ψ0=0 \implies {1 \over {\sqrt{2\hbar mw}} }(-ip+mwx) \psi_{0}=0

  ⟹  (−ip+mwx)ψ0=0 \implies (-ip+mwx) \psi_{0}=0

  ⟹  (ℏ∂∂x+mwx)ψ0=0 \implies (\hbar\frac{\partial}{\partial x} +mwx) \psi_{0}=0

  ⟹  ∂∂xψ0=−mwxℏψ0 \implies \frac{\partial}{\partial x}\psi_{0}=-\frac{mwx}{\hbar}\psi_{0}

By separating variables here,

1ψ0dψ0=−mwxℏdx \displaystyle \frac{1}{\psi_{0}} d\psi_{0}=-\frac{mwx}{\hbar} dx

  ⟹  ln⁡(ψ0)=−mwx22ℏ+C \implies \ln (\psi_{0}) = -\frac{mwx^{2}}{2\hbar}+C

  ⟹  ψ0(x)=Ce−mwx22ℏ \implies \psi_{0}(x)=Ce^{-\frac{mwx^{2}}{2\hbar}}

Now, we need to find the normalization constant CC to know the exact eigenfunction of the ground state. According to the normalization condition,

∫−∞∞(ψ0)∗ψ0dx=1 \displaystyle \int_{-\infty}^{\infty} (\psi_{0})^{\ast}\psi_{0}dx=1

  ⟹  ∣C∣2∫−∞∞e−mwx2ℏdx=1 \implies |C|^{2}\int_{-\infty}^{\infty}e^{-\frac{mwx^{2}}{\hbar}} dx=1

To evaluate this integral, refer to the Gaussian Integral: the standard integral of the form e−x2e^{-x^{2}}. By substituting for convenience,

mwℏx≡y \sqrt{\frac{mw}{\hbar}}x \equiv y

dx=ℏmwdydx=\sqrt{\frac{\hbar}{mw}}dy

The limits of integration remain unchanged. Substituting back into the original equation,

  ⟹  ∣C∣2ℏmw∫−∞∞e−y2dy=1 \displaystyle \implies |C|^{2}\sqrt{\dfrac{\hbar}{mw}}\int_{-\infty}^{\infty}e^{-^y2} dy=1

  ⟹  ∣C∣2ℏmwπ=1 \implies |C|^{2}\sqrt{\dfrac{\hbar}{mw}}\sqrt{\pi}=1

  ⟹  ∣C∣2ℏπmw=1 \implies |C|^{2}\sqrt{\dfrac{\hbar \pi}{mw}}=1

∴∣C∣2=mwℏπ,  C=(mwℏπ)14 \therefore |C|^{2}=\sqrt{\frac{mw}{\hbar \pi}},\ \ C=(\frac{mw}{\hbar \pi})^{\frac{1}{4}}

Thus, the final eigenfunction of the ground state is ψ0(x)=(mwℏπ)14e−mwx22ℏ\psi_{0} (x)=(\frac{mw}{\hbar \pi})^{\frac{1}{4}} e^{-\frac{mwx^{2}}{2\hbar}}. In the next post, we’ll find the generalized eigenfunctions for nn. Next post: Solving the Harmonic Oscillator Problem Using the Operator Method : Generalized Eigenfunctions.

Previous post: Energy Levels and Ground State Eigenfunctions of the Harmonic Oscillator.

Now, let’s use the ladder operators and the ground state eigenfunctions to find the generalized eigenfunctions of the harmonic oscillator. Ladder Operators a±a_\pm increase or decrease the state of an eigenfunction ψn\psi_{n} by one step. Thus, we can set the following equation.

$a_{+} \ket{\psi_{n}}=C_{+}|\psi_{n+1}>

a−∣ψn⟩=C−∣ψn−1⟩ a_{-} \ket{\psi_{n}}=C_{-} \ket{\psi_{n-1}}

C± C_\pm 는 각각 n n 번째와 (n+1) (n+1) 번째, n n 번째와 (n−1) (n-1) 번째 상태 사이의 비례계수이다. 이 비례계수를 정확하게 구해보자.각 고유함수들이 규격화된 고유함수라고 가정하고 규격화조건을 사용하자.규격화된 고유 함수는 자신과 내적하면 값이 11 이다.사다리 연산자와 해밀토니안 사이의 관계식을 사용하면.

(a+∣ψn⟩)∗(a+∣ψn⟩)=⟨ψn∣a−a+∣ψn⟩=⟨ψn∣1ℏwH+12∣ψn⟩=1ℏwEn<ψn∣ψn⟩+12<ψn∣ψn⟩=1ℏw(n+12)ℏw+12=n+1 \begin{align*} (a_{+} \ket{\psi_{n}})^{\ast}(a_{+} \ket{\psi_{n}}) &= \bra{\psi_{n}} a_{-}a_{+} \ket{\psi_{n}} \\ &= \bra{\psi_{n}} \frac{1}{\hbar w}H + \frac{1}{2} \ket{\psi_{n}} \\ &= \frac{1}{\hbar w} E_{n} <\psi_{n} \ket{\psi_{n}} + \frac{1}{2} <\psi_{n} \ket{\psi_{n}} \\ &= \frac{1}{\hbar w}(n+\frac{1}{2})\hbar w +\frac{1}{2} \\ &= n+1 \end{align*}

반면 위의 비례식을 사용하면

(a+∣ψn⟩)∗(a+∣ψn⟩)=(C+∣ψn+1>)∗(C+∣ψn+1>)=∣C+∣2⟨ψn+1∣ψn+1⟩=∣C+∣2 \begin{align*} (a_{+} \ket{\psi_{n}})^{\ast}(a_{+} \ket{\psi_{n}}) &= (C_{+}|\psi_{n+1}>)^{\ast}(C_{+}|\psi_{n+1}>) \\ &= |C_{+}|^{2}\braket{\psi_{n+1} | \psi_{n+1}} \\ &= |C_{+}|^{2} \end{align*}

따라서 위의 두 결과를 종합하면 C+ C_{+} 값을 얻을 수 있다.$ |C_{+}|^{2} = n+1.

  ⟹  C+=n+1 \implies C_{+}=\sqrt{n+1}

\therefore a_{+} \ket{\psi_{n}} = \sqrt{n+1}|\psi_{n+1}>같은방법으로 같은 방법으로 C_{-} 도구할수있다.과정은생략하고결과만적을테니직접해보길바란다.도 구할 수 있다.과정은 생략하고 결과만 적을 테니 직접 해보길 바란다. |C_{-}|^{2} = n.

  ⟹  C−=n \implies C_{-}=\sqrt n

  ⟹  a−∣ψn⟩=n∣ψn−1> \implies a_{-} \ket{\psi_{n}}=\sqrt n |\psi_{n-1}>

Now, using this result and the ground state ∣ψ0⟩ \ket{\psi_{0}}, let’s find the generalized nn state.

Since a+∣ψn⟩=n+1∣ψn+1> a_{+} \ket{\psi_{n}}=\sqrt{n+1}|\psi_{n+1}>, it is ∣ψn+1>=1n+1a+∣ψn⟩ |\psi_{n+1}>=\frac{1}{\sqrt{n+1}}a_{+} \ket{\psi_{n}}.

First excited state ∣ψ1⟩=a+∣ψ0⟩\ket{\psi_{1}}=a_{+} \ket{\psi_{0}}

Second excited state

∣ψ2>=12a+∣ψ1⟩=12a+a+∣ψ0⟩=12(a+)2∣ψ0⟩\begin{align*} |\psi_2> &= \frac{1}{\sqrt{2}}a_{+}\ket{\psi_{1}} \\ &= \frac{1}{\sqrt{2}}a_{+}a_{+} \ket{\psi_{0}} \\ &= \frac{1}{\sqrt{2}}(a_{+})^{2} \ket{\psi_{0}} \end{align*}

Third excited state

∣ψ3>=13a+∣ψ2>=13a+12(a+)2∣ψ0⟩=13!(a+)3∣ψ0⟩ \begin{align*} |\psi_{3}> &= \frac{1}{\sqrt{3}}a_{+}|\psi_2> \\ &= \frac{1}{\sqrt{3}}a_{+}\frac{1}{\sqrt{2}}(a_{+})^{2} \ket{\psi_{0}} \\ &= \frac{1}{\sqrt{3!}}(a_{+})^3 \ket{\psi_{0}} \end{align*}

Fourth excited state

∣ψ4>=14a+∣ψ3>=14a+13!(a+)3∣ψ0⟩=14!(a+)4∣ψ0⟩ \begin{align*} |\psi_{4}> &= \frac{1}{\sqrt{4}}a_{+}|\psi_{3}> \\ &= \frac{1}{\sqrt{4}}a_{+}\frac{1}{\sqrt{3!}}(a_{+})^3 \ket{\psi_{0}} \\ &= \frac{1}{\sqrt{4!}}(a_{+})^4 \ket{\psi_{0}} \end{align*}

Therefore, substituting the previously found eigenfunction of the ground state ψ0(x)=(mwℏπ)14e−mwx22ℏ\psi_{0}(x)=(\frac{mw}{\hbar \pi})^{\frac{1}{4}} e^{-\frac{mwx^{2}}{2\hbar}} and ladder operators a+=12ℏmw(mwx+ip)=mw2ℏx−ℏ2mwddxa_{+}=\frac{1}{\sqrt{2\hbar mw }}(mwx+ip)=\sqrt{\frac{mw}{2\hbar}}x-\sqrt{\frac{\hbar}{2mw}}\frac{d}{dx} into ψn(x)=1n!(mw2ℏx−ℏ2mwddx)n(mwℏπ)14e−mwx22ℏ\psi_{n} (x) =\frac{1}{\sqrt{n!}} \left( \sqrt{\frac{mw}{2\hbar}}x-\sqrt{\frac{\hbar}{2mw}}\frac{d}{dx} \right)^n (\frac{mw}{\hbar \pi})^{\frac{1}{4}} e^{-\frac{mwx^{2}}{2\hbar}}, this is the generalized nn state of the harmonic oscillator obtained using operators (the nn eigenfunction).