Field Quantization zero point energy

Taking up the eigenvalue equation for $\hat{H}(\hat{a}\vert n \rangle)$

\begin{equation} \hat{H} \left(\hat{a} \vert n \rangle \right)=  \left( E_n - \hbar\omega \right) \left( \hat{a} \vert n \rangle \right) \end{equation}

Applying the annihilation operator repeatedly the energy eigenvalue will decrease in integer multiples of $\hbar \omega$, the energy of harmonic oscillator must always be positive, therefore exists a lowest energy eigenvalue corresponding to eigenstate $\vert 0 \rangle$. The eigenvalue equation for the ground state is:

\begin{equation}\require{cancel} \hat{H}\vert 0 \rangle = \hbar\omega \left( \hat{a}^\dagger \hat{a} + \frac{1}{2} \right) \vert 0 \rangle = \cancelto{0}{ \hbar\omega \hat{n}\vert 0 \rangle} + \frac{1}{2}\hbar\omega \vert 0 \rangle = \frac{1}{2}\hbar\omega \vert 0 \rangle  \end{equation}

So the lowest energy eigenvalue is $E_0 = \cfrac{\hbar\omega}{2}$. Since $E_{n+1}= E_n + \hbar\omega$, the energy eigenvalues are:

\begin{equation}E_n = \hbar\omega \left(  n + \frac{1}{2} \right)  \end{equation}

Acknowledgement

I would like to thank Dr. Carlos Herman Wiechers Medina for their support, this post is based on the notes from the Quantum Optics class.

Read More

Field Quantization. Annihilation and Creation operators

Taking up the Hamiltonian equation from Field Quantization. Hamiltonian for a single-mode field

\begin{equation} H = \frac{1}{2} \left( p^2 + \omega^2 q^2  \right) \end{equation}

Where $q$ and $p$ are the canonical variables. To make a quantum approach  we just replace them with the respective operator $\hat{q}$ and $\hat{p}$, which must satisfy the canonical commutation relation:

\begin{equation} \left[ \hat{q},\hat{p} \right]=i\hbar \hat{I} \equiv i\hbar \end{equation}

With these operators the Hamiltonian transforms into the operator:

\begin{equation} \hat{H} = \frac{1}{2} \left( \hat{p} ^2 + \omega^2 \hat{q} ^2  \right)  \end{equation}

Here is convenient to introduce the non-Hermitian annihilation $\hat{a}$ and creation $\hat{a} ^\dagger$ operators given by:

\begin{equation} \hat{a} = (2 \hbar \omega)^{-1/2} (\omega \hat{q}+ i\hat{p}) \end{equation} \begin{equation} \hat{a} ^\dagger = (2 \hbar \omega)^{-1/2} (\omega \hat{q}- i\hat{p}) \end{equation}

These satisfies the commutation relation: 

\begin{equation}  \left[ \hat{a},\hat{a} ^\dagger \right] =  \hat{a} \hat{a} ^\dagger -  \hat{a}^\dagger \hat{a}= 1 \end{equation}

Noting that

\begin{align*} \hat{a} ^\dagger \hat{a} &= (2 \hbar \omega)^{-1/2} (\omega \hat{q}- i\hat{p})(2 \hbar \omega)^{-1/2} (\omega \hat{q}+ i\hat{p}) =  \frac{1}{2 \hbar \omega}\left[  \omega^2\hat{q}^2 + i\omega\hat{q}\hat{p}-i\omega\hat{p}\hat{q}+\hat{p}^2  \right] \\ \\ &=  \frac{1}{2 \hbar \omega}\left[  \omega^2\hat{q}^2 + i\omega \left[ \hat{q},\hat{p} \right]+\hat{p}^2  \right] =  \frac{1}{2 \hbar \omega}\left[  \omega^2\hat{q}^2 - \omega\hbar +\hat{p}^2  \right] = \frac{1}{\hbar \omega} \frac{1}{2} \left( \hat{p}^2 +  \omega^2\hat{q}^2 \right)  - \frac{1}{2} \\ \\ &= \frac{1}{\hbar \omega}\hat{H}-\frac{1}{2} \end{align*}

So, the Hamiltonian operator can be defined as:

\begin{equation} \hat{H} = \hbar\omega \left(  \hat{a} ^\dagger \hat{a} + \frac{1}{2} \right) \end{equation}

Now we can describe the energy eigenvalues. We denote $\vert n \rangle$ as energy eigenstate of the single mode field with eigenvalue $E_n$ such that:

\begin{equation} \hat{H}\vert n \rangle =  \hbar\omega \left(  \hat{a} ^\dagger \hat{a} + \frac{1}{2} \right) \vert n \rangle = E_n \vert n \rangle \end{equation}

The operator $\hat{a}^\dagger \hat{a}$ has a great importance and is called number operator $\hat{n}$. We will generate the eigenvalue equations for $\hat{a}^\dagger \vert n \rangle$ and $\hat{a} \vert n \rangle$ to appreciate the effect of the operators in the energy level and understand why they are called that. Multiplying Eq. (8) by  $\hat{a}^\dagger$ we obtain:

\begin{equation*} \hbar\omega \left( \hat{a} ^\dagger \hat{a} ^\dagger \hat{a} + \frac{1}{2}\hat{a} ^\dagger \right) \vert n \rangle =  E_n \hat{a}^\dagger \vert n \rangle \end{equation*}

\begin{equation*} \hbar\omega \left( \hat{a} ^\dagger \left(\hat{a}\hat{a}^\dagger -1 \right) + \frac{1}{2}\hat{a} ^\dagger \right) \vert n \rangle =  E_n \hat{a}^\dagger \vert n \rangle \end{equation*}

\begin{equation*} \hbar\omega \left( \hat{a} ^\dagger\hat{a}\hat{a}^\dagger + \frac{1}{2}\hat{a} ^\dagger \right) \vert n \rangle =   E_n  \hat{a}^\dagger \vert n \rangle +  \hbar\omega  \hat{a}^\dagger \vert n \rangle \end{equation*}

\begin{equation*} \hbar\omega \left( \hat{a} ^\dagger\hat{a} + \frac{1}{2} \right) \left(\hat{a}^\dagger \vert n \rangle \right)=  \left( E_n + \hbar\omega \right) \left( \hat{a}^\dagger \vert n \rangle \right) \end{equation*}

\begin{equation} \Rightarrow \hat{H} \left(\hat{a}^\dagger \vert n \rangle \right)=  \left( E_n + \hbar\omega \right) \left( \hat{a}^\dagger \vert n \rangle \right) \end{equation}

This is the equation for the eigenstate  $\hat{a}^\dagger \vert n \rangle$, it has energy eigenvalue of $E_n + \hbar \omega$. We can interpret this as if the operator $\hat{a}^\dagger$ created one quantum of energy $\hbar \omega$, that is why it is called as the creation operator. In the same way, we can multiply Eq. (8) by $\hat{a}$ to obtain the equation for the eigenstate $\hat{a}\vert n \rangle$:

\begin{equation} \hat{H} \left(\hat{a} \vert n \rangle \right)=  \left( E_n - \hbar\omega \right) \left( \hat{a} \vert n \rangle \right) \end{equation}

Analogous to the previous case, we can interpret it as if the operator $\hat{a}$ eliminated or annihilated one quantum of energy $\hbar \omega$, that is why it is called as the annihilation operator. 

For the states $\hat{a}\vert n \rangle$, $\hat{a}^\dagger \vert n \rangle$ and $\hat{n} \vert n \rangle$ we have:

\begin{equation}\hat{a}\vert n \rangle = c_n \vert n-1 \rangle \end{equation}

\begin{equation}\hat{a}^\dagger \vert n \rangle = d_n \vert n+1 \rangle \end{equation}

\begin{equation}\hat{n} \vert n \rangle = n \vert n \rangle \end{equation}

Where $c_n$ and $d_n$ are constants. The number of states must be normalized, i.e. $\langle n \vert n \rangle = 1$. The inner product of  $\hat{a}\vert n \rangle$ with itself is:

\begin{equation} (\langle n \vert \hat{a}^\dagger)(\hat{a}\vert n \rangle) = \langle n \vert \hat{a}^\dagger \hat{a} \vert n \rangle = \langle n \vert \hat{n} \vert n \rangle = n \langle n \vert n \rangle = n \end{equation}

Also

\begin{equation} (\langle n \vert \hat{a}^\dagger)(\hat{a}\vert n \rangle) = \langle n-1 \vert c_n^* c_n  \vert n-1 \rangle = \vert c_n \vert ^2 \langle n-1 \vert n-1 \rangle = \vert c_n \vert ^2 \end{equation}

Thus $c_n = \sqrt{n}$ and

\begin{equation} \boxed{ \hat{a}\vert n \rangle = \sqrt{n} \vert n-1 \rangle} \end{equation}

In the same way, the inner product of  $\hat{a}^\dagger \vert n \rangle$ with itself is:

\begin{equation} (\langle n \vert \hat{a})(\hat{a}^\dagger \vert n \rangle) = \langle n \vert \hat{a} \hat{a}^\dagger \vert n \rangle = \langle n \vert (\hat{a}^\dagger\hat{a}+1) \vert n \rangle = \langle n \vert \hat{n} \vert n \rangle +  \langle n \vert  n \rangle =  n + 1\end{equation}

Also

\begin{equation} (\langle n \vert \hat{a})(\hat{a}^\dagger \vert n \rangle) = \langle n+1 \vert d_n^* d_n \vert n+1 \rangle = \vert d_n \vert ^2 \langle n+1 \vert n+1 \rangle = \vert d_n \vert ^2 \end{equation}

Thus $d_n = \sqrt{n+1}$ and

\begin{equation}\boxed{ \hat{a}^\dagger \vert n \rangle = \sqrt{n+1} \vert n+1 \rangle} \end{equation}

Acknowledgement

I would like to thank Dr. Carlos Herman Wiechers Medina for their support, this post is based on the notes from the Quantum Optics class.

Read More

Field Quantization. Hamiltonian for a single-mode field

We have the case of a radiant field confined in one dimensional cavity along the z-axis with perfectly conducting walls at  $z = 0$ and $z = L$, so that the electric field vanish in the boundaries as shown in:

Maxwell's equations without sources are:

\begin{equation} \nabla \times \textbf{E} = \frac{\partial \textbf{B}}{\partial t}  \end{equation}

\begin{equation} \nabla \times \textbf{B} = \mu_0 \varepsilon_0 \frac{\partial \textbf{E}}{\partial t}  \end{equation}

\begin{equation} \nabla \cdot \textbf{B} = 0  \end{equation}

\begin{equation}  \nabla \cdot \textbf{E} = 0  \end{equation}

The field is polarized in x-direction, i.e. $\textbf{E}(\textbf{r},t) = \mathbf{e}_x  E_x(z,t)$, with $\textbf{e}_x $ an unit polarization vector. A field that satisfies Maxwell's equations and boundary conditions is:

\begin{equation} E_x(z,t)=\left( \frac{2\omega^{2}}{V\varepsilon_0} \right)^{1/2}q(t) \text{sin}(kz)  \end{equation}

With $\omega$ the frequency of the mode, $k=\cfrac{\omega}{c}$ the wave number, $V$ the effective volume of the cavity and $q(t)$ the canonical position. We can not have all frequencies due to the boundary condition at $z=L$, this only allowed frequencies $\omega_m = c \cfrac{m\pi}{L}, \quad m=1,2,\cdots$. From Eq. (2) and Eq. (5) we obtain: 

\begin{equation*} \textbf{e}_x \left( \partial_y B_z - \partial_z B_y  \right) + \textbf{e}_y \left( \partial_z B_x - \partial_x B_z  \right) + \textbf{e}_z \left( \partial_x B_y - \partial_y B_x  \right) = \textbf{e}_x \left( \mu_0\varepsilon_0 \partial_t E_x \right) \\ \partial_y B_z - \partial_z B_y = \mu_0\varepsilon_0 \left( \frac{2\omega^{2}}{V\varepsilon_0} \right)^{1/2} \text{sin}(kz) \partial_t q =  \mu_0\varepsilon_0 \left( \frac{2\omega^{2}}{V\varepsilon_0} \right)^{1/2} \dot{q}(t) \text{sin}(kz)  \\  - \partial_z B_y = \mu_0\varepsilon_0 \left( \frac{2\omega^{2}}{V\varepsilon_0} \right)^{1/2} \dot{q}(t) \text{sin}(kz) \end{equation*}

\begin{equation} \Rightarrow  B_y (z,t) = \frac{\mu_0\varepsilon_0}{k} \left( \frac{2\omega^{2}}{V\varepsilon_0} \right)^{1/2} \dot{q}(t) \text{cos}(kz) \end{equation}

The Hamiltonian of the field described above is:

\begin{equation} H=\frac{1}{2}\int dV \left[ \varepsilon_0 E_x^{2}(z,t) + \frac{1}{\mu_0}B_y^{2}(z,t) \right] \end{equation}

Substituting Eq. (5) and Eq. (6) in Eq. (7):

\begin{equation*} H=\frac{1}{2} \left[  \varepsilon_0 \frac{2\omega^{2}}{V\varepsilon_0} q^2(t) \int dV \text{sin}^2(kz)  + \frac{1}{\mu_0} \frac{\mu_0^2 \varepsilon_0^2}{k^2} \frac{2\omega^{2}}{V\varepsilon_0} \dot{q}^2(t) \int dV \text{cos}^2(kz) \right] \end{equation*}

It is important to emphasize that $V$ in the equations for $E_x$ and $B_y$ is the volume of the cavity, i.e. it is a constant. Noting that $k=\cfrac{\omega}{c}$ and $\mu_0\varepsilon_0 = \cfrac{1}{c^2}$

\begin{equation*} H=\frac{1}{2} \left[  \frac{2\omega^{2}}{V} q^2(t) \int dV \text{sin}^2(kz)  +   \frac{2}{V} \dot{q}^2(t) \int dV \text{cos}^2(kz) \right] \end{equation*}

For this case we have $\int dV = A\int_L dz$, so:

\begin{align*} H &= \frac{1}{2}A \left[  \frac{2\omega^{2}}{V} q^2(t) \int_L dz \; \text{sin}^2(kz)  +   \frac{2}{V} \dot{q}^2(t) \int_L dz \; \text{cos}^2(kz) \right] \\ \\ &= \frac{1}{2}A \left[ \frac{\omega^2 q^2(t)L}{V} +  \frac{\dot{q}^2(t) L}{V}  \right]\end{align*}

Considering a particle of unit mass, i.e. the canonical momentum is $p(t) = \dot{q}(t)$ and noting that $V=AL$, the Hamiltonian results:

\begin{equation} \boxed{H = \frac{1}{2} \left( p^2 + \omega^2 q^2  \right)} \end{equation}

Acknowledgement

I would like to thank Dr. Carlos Herman Wiechers Medina for their support, this post is based on the notes from the Quantum Optics class.

Read More

Dirac's Notation

In quantum mechanics, we can describe "pure" quantum states using an n-dimensional vector or rank 1 tensor, this is called ket and is represented with: $$  \vert \Psi  \rangle  =  \begin{pmatrix} \alpha_1 \\ \vdots \\ \alpha_n \end{pmatrix} $$

We can form the corresponding bra (dual space) transposing the ket and complex conjugating every component: $$  \langle \Psi  \vert  =  \begin{pmatrix} \alpha_1^* & \cdots & \alpha_n^*  \end{pmatrix} $$ At first instance we can define some products using bras and kets, the inner product, outer product and tensorial product.

Inner product:  \begin{equation*} \langle \Psi  \vert \Psi  \rangle  =  \begin{pmatrix} \alpha_1^* & \cdots & \alpha_n^* \end{pmatrix} \begin{pmatrix} \alpha_1 \\ \vdots \\ \alpha_n \end{pmatrix} =  \alpha_1^*  \alpha_1 + \cdots \alpha_n^*  \alpha_n =  \vert \alpha_1 \vert ^2 + \cdots + \vert \alpha_n \vert ^2  \end{equation*}   

Outer product: \begin{align*} \vert  \Psi \rangle \langle  \Psi  \vert  &=   \begin{pmatrix} \alpha_1 \\ \vdots \\ \alpha_n \end{pmatrix} \begin{pmatrix} \alpha_1^* & \cdots & \alpha_n^* \end{pmatrix} =  \begin{pmatrix}  \alpha_1  \alpha_1^* &  \alpha_1  \alpha_2^*  & \cdots &  \alpha_1  \alpha_n^* \\  \alpha_2  \alpha_1^* &  \alpha_2  \alpha_2^*  & \cdots &  \alpha_2  \alpha_n^* \\ \vdots & \vdots & \ddots & \vdots \\  \alpha_n  \alpha_1^* &  \alpha_n  \alpha_2^*  & \cdots &  \alpha_n  \alpha_n^* \end{pmatrix} \\ \\  &=   \begin{pmatrix}  \vert \alpha_1 \vert ^2 &  \alpha_1  \alpha_2^*  & \cdots &  \alpha_1  \alpha_n^* \\  \alpha_2  \alpha_1^* &  \vert \alpha_2 \vert ^2  & \cdots &  \alpha_2  \alpha_n^* \\ \vdots & \vdots & \ddots & \vdots \\  \alpha_n  \alpha_1^* &  \alpha_n  \alpha_2^*  & \cdots & \vert \alpha_n \vert ^2  \end{pmatrix} \end{align*}

Tensorial product: $ \vert \alpha \rangle \otimes \vert \beta \rangle \equiv \vert \alpha \rangle \vert \beta \rangle \equiv  \vert \alpha , \beta \rangle \equiv  \vert \alpha \beta \rangle $. Supposing 2-dimensional kets $ \vert \alpha \rangle = \big(\begin{smallmatrix} \alpha_1 \\ \alpha_2 \end{smallmatrix}\big) $ and $\vert \beta \rangle = \big(\begin{smallmatrix} \beta_1 \\ \beta_2 \end{smallmatrix}\big) $ we can define the tensorial product as:

\begin{align*} \vert \alpha \rangle \otimes \vert \beta \rangle  &=   \begin{pmatrix} \alpha_1 \\ \alpha_2 \end{pmatrix} \otimes \begin{pmatrix} \beta_1 \\ \beta_2 \end{pmatrix} =  \begin{pmatrix} \alpha_1 \begin{pmatrix} \beta_1 \\ \beta_2  \end{pmatrix} \\ \alpha_2  \begin{pmatrix} \beta_1 \\ \beta_2  \end{pmatrix} \end{pmatrix} = \begin{pmatrix} \alpha_1 \beta_1 \\  \alpha_1 \beta_2  \\  \alpha_2 \beta_1 \\  \alpha_2 \beta_2  \end{pmatrix}  \end{align*} 

Read More

About Myself

Hello my name is David Angel Alba Bonilla, a physics engineering student in Universidad de Guanajuato. I like to know how the nature and the technology works using physics, mathematics and science in general. I want to expose my knowledge, my reasonings, and my love of the physics with everybody I can. You may think physics is so difficult, it's useless in daily life, or the physics is only "use formulas",  I will explain why all of these ideas are wrong on this blog.

I hope you enjoy the blog and the information be useful for all of you.

Read More