I wanted to rigorously understand the derivation of the ground state (of the quantum harmonic oscillator) using the position representation $\Psi_0(x)=\langle x|0\rangle$ and $a|0\rangle=0$. There I have:

\begin{align*}0=a|0\rangle\implies 0&=\frac{1}{2x_0} \langle x|\left(x+\frac{i}{m\omega}\hat{p} \right)|0\rangle=\frac{1}{2x_0}\left(x+\frac{i}{m\omega}\hat{p}\right)\langle x|0\rangle \\ &=\frac{x}{2x_0}\psi_0(x)+\frac{\hbar}{2x_0m\omega}\frac{d}{dx}\psi_0(x), \end{align*}

where $\hat{p}=-i\hbar\frac{d}{dx}$ and $x_0=\sqrt{\frac{\hbar}{2m\omega}}$.

The step that really confuses me here is that we seemingly use $$\langle x| \hat{p} |0\rangle = \hat{p}\langle x|0\rangle$$ where I fail to see why this holds true.

Any help would be greatly appreciated.

  • 1
    $\begingroup$ Your last equation cannot be true: on the LHS, there is a $\mathbb C$-number, while the RHS is an operator (times a $\mathbb C$ number). However, I guess that's a quite common abuse of notation. Rather, it should be $\langle x|\hat p = -i \hbar \frac{\mathrm d}{\mathrm d x} \langle x|$, such that $\langle x|\hat p |0\rangle = -i\hbar \frac{\mathrm d}{\mathrm d x} \langle x|0\rangle$. $\endgroup$ Commented Jan 2, 2022 at 22:49
  • $\begingroup$ Related: physics.stackexchange.com/q/76299/2451 and links therein. $\endgroup$
    – Qmechanic
    Commented Jan 3, 2022 at 11:59

3 Answers 3


$\renewcommand\mean[1]{\langle #1\rangle}$ $\renewcommand\norm[1]{||#1||}$ $\renewcommand\h{\hbar}$ $\renewcommand\ket[1]{|#1\rangle}$ $\renewcommand\expval[2]{\langle #1|#2|#1\rangle}$ $\renewcommand\braket[2]{\langle #1|#2\rangle}$ $\renewcommand\Braket[3]{\langle #1|#2|#3\rangle}$ OP's original question

The step that really confuses me here is that we seemingly use \begin{equation} \Braket{x}{p}{0}=p\braket{x}{0}\end{equation} where I fail to see why this holds true.

has already been answered in the replies of @Cosmas and @Zack. I won't go over these again. Instead, I propose here an alternative method of finding the ground state of the Harmonic Oscillator, using this time the concept of coherent states. I am going to assume the reader knows nothing about them, so I will build up everything here to avoid losing anyone.

Introduction to coherent states

There are three different - equivalent - ways to define a coherent state. For the sake of answering your question, I will only need one of them (sadly the less intuitive of them all) so I won't go all the way to introduce them all. The plan is thus the following: defining the so-called Heisenberg Uncertainty Principle and using it to define a coherent state. From there, I will be able to build a mathematical structure that will allow us to find the ground state of the Quantum Harmonic Oscillator.

Theorem 1 (Heisenberg Uncertainty Principle) For any operators $\hat{A}$ and $\hat{B}$, we have that $\Delta\hat{A}\hat{B}\geq\frac{1}{2}\norm{\mean{[\hat{A},\hat{B}]}}$.

In the case of the position and momentum operators, one has that $[\hat{X},\hat{P}] = i\h$. Applying this to the uncertainty principle, we find the famous relation $\Delta\hat{X}\Delta\hat{P}\geq \frac{\h}{2}$.

Definition 2 (Coherent states) A coherent state is defined as a state for which the Heisenberg Uncertainty Principle (applied to the operators $\hat{X}$ and $\hat{P}$ in the context of your question) is saturated, such that $\Delta\hat{X}\Delta\hat{P}=\frac{\h}{2}$.

Wave function of a coherent state

My goal here is to construct a wave function that depicts how a coherent state behaves. Let us introduce $\mathcal{O}(\lambda)=f^\dagger(\lambda)f(\lambda)$, where we define $f(\lambda) = \hat{A'}+i\lambda\hat{B'}, \hat{A'} = \hat{A}-\mean{\hat{A}}$ and $\hat{B'} = \hat{B}-\mean{\hat{B}}$.

Let us observe that the mean of $\mathcal{O}$ is positive. Indeed, for any $\ket{\psi}\in\mathcal{H}$, one has that \begin{equation} \expval{\psi}{f^\dagger(\lambda)f(\lambda)} = \norm{f(\lambda)\ket{\psi}}^2 \geq 0 \end{equation}

However, one also has that \begin{equation}\tag{1}\label{mean value O} \mean{\mathcal{O}(\lambda)} = \lambda^2\mean{\hat{B'}}+\lambda\mean{i[\hat{A'},\hat{B'}]}+\mean{\hat{A'}^2}. \end{equation} Notice that this is a quadratic equation in $\lambda$. We can compute its minimum through derivation. One finds that $\lambda_{min} = -\frac{\mean{i[\hat{A'},\hat{B'}]}}{2\left(\Delta\hat{B'}\right)^2}$. Notice that it corresponds indeed to a minimum, for the coefficient in front of $\lambda^2$ is positive in \eqref{mean value O} (which is basically a parabola).

We now apply our findings to $\hat{A}=\hat{X}$ and $\hat{B}=\hat{P}$. The minimum is then $\frac{\h}{2\left(\Delta\hat{P}\right)^2}$. In particular, one has that $\mean{\mathcal{O}(\lambda)} = 0$ if and only if \begin{align} \left(\hat{X'}+i\lambda_{min}\hat{P'}\right)\ket{\psi} &= 0\\ \left(X-\mean{X}\right)\ket{\psi} &= -\frac{i\h}{2\left(\Delta\hat{P}\right)^2}\left(\hat{P}-\mean{\hat{P}}\right)\ket{\psi} \end{align} Let us now write $l = \Delta\hat{X}$. Because we define a coherent state as the saturation of the Heisenberg uncertainty principle, we can write $2\left(\Delta\hat{P}\right)^2 = \frac{\h^2}{l^2}$. Hence, one has that \begin{equation} \left(\hat{X}-\mean{\hat{X}}\right)\ket{\psi} = -i\frac{l^2}{\h}\left(\hat{P}-\mean{\hat{P}}\right)\ket{\psi}\tag{2}\label{b wave function} \end{equation} Using the x-representation expression of $\hat{P}$ (see answers above for more details on this), we find the differential equation \begin{align} -x\psi(x)+\expval{\hat{X}}\psi(x)&=\frac{il^2}{\h}\left(\frac{\h}{i}\psi'(x)-\expval{\hat{P}}\psi(x)\right)\notag\\ \psi'(x) &= \left[\frac{\mean{\hat{X}}-x}{l^2}+\frac{i}{\h}\mean{\hat{P}}\right]\psi(x)\tag{3}\label{wave function} \end{align} The solution to this is given by $\psi(x) = \psi_0\exp(-\frac{\left(x-\mean{\hat{X}}\right)^2}{2l^2}+\frac{i}{\h}\mean{\hat{P}}x)$ where $\psi_0$ is a normalization constant.

Through an appropriate Gaussian integral, one finds that the wave function of a coherent state is given by \begin{equation} \psi(x) = \frac{\exp(-\frac{i}{2\pi}\mean{\hat{X}}\mean{\hat{P}})}{\left(\pi l^2\right)^{1/4}}\exp\left(-\frac{\left(x-\mean{\hat{X}}\right)^2}{2l^2}+\frac{i\mean{\hat{P}}x}{\hbar}\right)\tag{4}\label{wave function coherent states} \end{equation} \eqref{wave function coherent states} shows that the sought after wave function is a gaussian. Notice we added a global phase of $\exp(-\frac{i}{2\pi}\mean{\hat{X}}\mean{\hat{P}})$. Global phases are arbitrary and do not change the physics of the problem. An appropriate phase might be chosen in such a way as to simplify future considerations, and that is what is done here.

Interlude : other formulations of coherent states

I am going to develop here the mathematics behind coherent states, so as to find two other formulations of them that are easier to use. In particular, we're going to be able to use algebra instead of calculus - which is arguably (in my experience) simpler to manipulate. The reader only interested in the proposed alternative solution to the problem arised by the OP may skip to the next section.

Eigenvectors of the annihilation operator

Let us introduce $\hat{a} = \frac{1}{2}\left(\frac{\hat{X}}{l}+\frac{il\hat{P}}{\h}\right)$. One easily checks that $[\hat{a},\hat{a}^\dagger] = \mathbb{I}$. We call $\hat{a}$ the annihilation operator, whereas $\hat{a}^\dagger$ is known as the creation operator. The reasons behind these names are obvious when one studies the eigenvalues of $\hat{N} = \hat{a}^\dagger\hat{a}$.

Let $\ket{\psi}$ be an element of an Hilbert space $\mathcal{H}$. What are the solutions of the eigen-value problem $\hat{a}\ket{\psi} = \alpha\ket{\psi}$? Using the defintion of $\hat{a}$ provided, one has that $\frac{1}{2}\left(\frac{\hat{X}}{l}+\frac{il\hat{P}}{\h}\right)\ket{\psi} = \alpha\ket{\psi}$. One steadily finds that $\alpha = \frac{1}{2}\left(\frac{\mean{\hat{X}}}{l}+\frac{il\mean{\hat{P}}}{\h}\right)$. Putting this in the eigen-value problem, one finds (after a few appropriate manipulations) the equation \eqref{b wave function} again.

This means that the wave function defined by \eqref{wave function coherent states} corresponds to the eigenvectors of $\hat{a}$. Per convention, we write $\ket{\psi} = \ket{\alpha}$.

Theorem 3 $\ket{\alpha}$ is a coherent state (as defined in Definition 2) if and only if $\ket{\alpha}$ verifies \begin{equation} \hat{a}\ket{\alpha}=\alpha\ket{\alpha}\tag{5}\label{coherent in eigen} \end{equation} for any $\alpha\in\mathbb{C}$.

Displacement operator

We define the displacement operator as $\hat{D}(\alpha) = e^{\alpha\hat{a}^\dagger-\alpha^*\hat{a}}$.

Property 4 (Glauber) Let $\hat{A}$ and $\hat{B}$ be two operators. The relation $e^\hat{A}e^\hat{B} = e^{\hat{A}+\hat{B}}e^{-\frac{1}{2}[\hat{A},\hat{B}]}$ holds true if and only if ($\hat{A},\hat{B}$) commutes with [$\hat{A},\hat{B}$].

Through an appropriate use of Property 4, one shows the displacement operator defined above can equivalently be defined as $\hat{D}(\alpha) = e^{-\frac{1}{2}\mathbb{I}}e^{\alpha\hat{a}^\dagger}e^{-\alpha^*\hat{a}}$.

Property 5 One can show that a coherent state can be defined as the Displacement operator applied to the ground state $\ket{0}$: \begin{equation} \ket{\alpha} = \hat{D}(\alpha)\ket{0}\tag{6}\label{eq:coherent in displacement} \end{equation}

Here you go! Three equivalent formulations of coherent states. Of course, this was a very quick treatment and the reader interested in further developments in the topic may consult other sources. I suggest reading Prof. Reinhold's lecture notes on the matter.

Alternative solution to the question

Without any further mathematical developments or proofs, let us bring to the reader's attention (I may edit the post, later on, to add some maths to back up the following claim) that coherent states closely imitates the classical solutions of the harmonic oscillator. That means that if applied to the Hamiltonian of an harmonic oscillator, the means values of the position and impulsion operators follow the classical solution of the Harmonic Oscillator. The reader may read more about this here.

The OP did not specify, but I reckon he used the ground state in the Fock space $\ket{n}$. Although coherent states and Fock states are not the same, it turns out their ground states are equivalent. Moreover, one may prove the following relation \begin{equation} \ket{\alpha} = e^{-\frac{1}{2}\norm{\alpha}^2}\sum_n \frac{\alpha^n}{\sqrt{n!}}\ket{n} \end{equation} which would allow one to define a coherent state as a linear combination of Fock states. Using this, one may see that the ground state of a quantum harmonic oscillator is $\braket{x}{0} = \psi_0(x)$. Taking $\ket{0}$ to be the coherent void, one only has to put $\alpha = 0$ in \eqref{wave function coherent states}. This gives us an alternative way of computing of the ground state of the quantum harmonic oscillator!

PS: I don't have time to read everything I wrote now. There is probably lots of grammatical mistakes, and I hope as little as possible mathematical ones. Feel free to edit if it feels necessary.

  • $\begingroup$ I corrected a few details, added a reference to this post and added an equation in the last part. $\endgroup$
    – Juian
    Commented Jan 4, 2022 at 11:08

Just to elaborate on Cosmas's answer: when one writes $\hat{p} = -i \hbar \ d/dx$, what they really mean is that $\hat{p}$ is given by this expression in the $x$-representation. In other words, $\hat{p}$ has the following matrix elements with respect to the $|x\rangle$ basis: $$ \langle x | \hat{p} | x' \rangle = -i \hbar \frac{d}{dx} \delta(x-x') $$ One way you can "prove" the above is to insert a resolution of the identity $\int dp |p \rangle \langle p|$, use $\hat{p} |p \rangle = p |p\rangle$, and $\langle x | p \rangle = e^{ipx/\hbar} / \sqrt{2\pi \hbar}$. (I say "prove" in quotes, because it's somewhat a matter of what you take as your starting point: the definition of the operator, or the overlap $ \langle x | p \rangle$.)

Then, for any state $|\psi \rangle$ with position space wavefunction $\langle x | \psi \rangle = \psi(x)$, $$ \langle x | \hat{p} |\psi \rangle = \int dx' \langle x | \hat{p} | x' \rangle \langle x' | \psi \rangle = \int dx' \left( -i \hbar \frac{d}{dx} \delta(x-x') \right) \psi(x') = -i \hbar \frac{d}{dx} \psi(x) $$


You are misunderstanding the Dirac notation language. The proper relation you write should be $$\langle x| \hat{p} |0\rangle = \hat{p}_x \langle x|0\rangle, $$ i.e., the right-hand side operator is the realization of the momentum operator acting on functions of x, not vectors (kets), as it does on the left-hand side. It is shorthand for the less confusing relation $$ \hat p = \int\!\! dx ~ |x\rangle (-i\hbar\partial_x) \langle x| = \int\!\! dp~ |p\rangle p \langle p|, $$ as your instructor probably covered. Try sandwiching this between $\langle x'|$ and $|0\rangle$.



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