# Energy basis to the X basis

On Shankar page 217 when going from the operator representation to the differential representation he starts with

$$a|0\rangle = 0$$

And says that with a projection on the X basis we get

$$|0\rangle \rightarrow \langle x|0\rangle = \psi_0(x)$$

And with $$a=\frac{1}{\sqrt{2}}\left(y+\frac{d}{dy}\right)$$

It’s stated that in the X basis of our first equation ($$a|0\rangle=0$$) we get $$\left(y+\frac{d}{dy}\right)\psi_0(y)=0$$

I’m wondering how this makes any sense. From the way it’s set up it sounds like he’s doing $$\langle x|a|0\rangle$$ which would be annihilated as per the first equation. But the result he gets is the same as if he did $$a \langle x|0\rangle$$ but you can’t just pull the $$a$$ out because it has a $$p$$ in its definition. What is going on here?

• Hint: $\langle x | 0 \rangle$ is a number. – DanielSank Nov 2 '18 at 23:39
• Checked WP? – Cosmas Zachos Nov 3 '18 at 15:05

First, we need to lay out some preliminaries in the language of Shankar's book. If you want, you can skip to the bottom where I apply the following to your question.

When you express a state $$|\psi\rangle$$ in the position basis, what you're doing is applying the identity operator $$\mathbb{1} = \int dx \ |x\rangle\langle x|$$ and defining $$\langle x | \psi \rangle \equiv \psi(x)$$ to be the position-space wave function corresponding to the state $$|\psi\rangle$$.

$$|\psi \rangle = \int dx \ |x \rangle\langle x | \psi \rangle \equiv \underbrace{\int dx \ |x \rangle \underbrace{\psi(x)}_{\text{Wavefunction}}}_{\text{Entire state}}$$

When you express an operator $$\hat O$$ in the position basis, what you're doing is applying the identity operator $$\mathbb{1} = \int dx |x \rangle\langle x |$$ from the right and $$\mathbb{1} = \int dx' |x' \rangle\langle x' |$$ from the left and defining $$\langle x'|\hat O|x\rangle \equiv O_{xx'}$$ to be the position-space representation of the operator $$\hat O$$.

$$\hat O = \iint dx \ dx'\ |x'\rangle\langle x'|\hat O |x \rangle \langle x | \equiv \iint dx\ dx' \ |x'\rangle O_{xx'} \langle x|$$

( Note that $$\hat O$$ eats states and spits out other states, while $$O_{xx'}$$ eats (wave)functions and spits out other functions.)

From here, we can write, we can write

$$\hat O |\psi\rangle = \left(\iint dx\ dx' |x'\rangle O_{xx'} \langle x| \right)\left(\int dy \ |y\rangle \psi(y)\right)$$ $$= \iiint dx \ dx' \ dy \ |x'\rangle O_{xx'} \langle x|y\rangle \psi(y)$$

We can use the fact that $$\langle y|x\rangle = \delta(x-y)$$ to eliminate the integration over $$y$$ and yield

$$\hat O | \psi\rangle = \iint dx \ dx' |x'\rangle O_{xx'} \psi(x)$$

This is the most general possible expression for action of the operator $$\hat O$$ on the state $$|\psi\rangle$$ as expressed in the position basis.

In your question, you refer to $$\hat a \equiv \frac{\hat X + i\hat P}{\sqrt{2}}$$ (where we've set all the constants equal to 1 for simplicity - you can put them back in as an exercise). In the position basis, we have that $$X_{xx'} \equiv \langle x' | \hat X | x\rangle = x \delta(x-x')$$ and $$P_{xx'} \equiv \langle x' | \hat P | x\rangle = -i\delta'(x-x') = -i \delta(x-x') \frac{d}{dx'}$$ and so

$$a_{xx'}\equiv \langle x' | \hat a | x\rangle = \delta(x-x') \left(\frac{x + \frac{d}{dx'}}{\sqrt{2}}\right)$$

The wavefunction of your state is unknown at the moment, but you can call it $$\psi_0(x)$$. Plugging the operator $$a_{xx'}$$ into the integral we found earlier yields

$$\hat a|0\rangle = \iint dx \ dx' |x'\rangle \delta(x-x') \left(\frac{x + \frac{d}{dx'}}{\sqrt{2}}\right) \psi_0(x) = \int dx |x\rangle \frac{1}{\sqrt{2}} \left( x + \frac{d}{dx}\right)\psi_0(x)$$

which means that $$\hat a|0\rangle$$ yields a new state whose wavefunction is given by

$$\phi(x) = \frac{1}{\sqrt{2}}\left(x + \frac{d}{dx}\right) \psi_0(x) = 0$$

where the last equality arises because we know that $$\hat a$$ annihilates the state $$|0\rangle$$.

Your confusion comes from the fact that the annihilation operator $$\hat a$$ is in fact not equal to $$\frac 1{\sqrt{2}}(x+\frac d{dx})$$. $$\hat a$$ is an operator acting on an abstract Hilbert space in which the state vector $$|\psi \rangle \in \mathcal H$$ exists. The differential operator $$\frac 1{\sqrt{2}}(x+\frac d{dx})$$ is an operator acting on the space of functions, not the abstract vectors of $$\mathcal H$$. It is the action of $$a$$ on any vector $$|\psi \rangle$$ in the position representation ($$x$$ eigenbasis) that resembles the differential operation $$\frac 1{\sqrt{2}}(x+\frac d{dx})$$.

To see this more clearly, note that the state vector $$|\psi \rangle$$ is not even a function of $$x$$, whereas the wavefunction $$\psi(x) \equiv \langle x | \psi \rangle$$ is.

More explicitly: $$\hat{a}|\psi \rangle \neq \frac 1{\sqrt{2}}(x+\frac d{dx}) |\psi \rangle \qquad \text{doesn't even make sense since |\psi \rangle is not a function of x}$$

What does make sense is decomposing $$|\psi \rangle$$ in the $$x$$ eigenbasis as $$|\psi \rangle \equiv \int_{-\infty}^\infty dx |x \rangle \langle x|\psi \rangle$$ or equivalently $$|\psi \rangle =\int_{-\infty}^\infty dx |x \rangle \psi(x)$$, and then saying that: $$\hat{a}|\psi \rangle = \int_{-\infty}^\infty dx \frac 1{\sqrt{2}}(x \psi(x)+\frac d{dx}\psi(x))|x \rangle \qquad \text{makes sense, both sides are vectors in \mathcal H}$$ Or equivalently, by multiplying $$\langle x|$$ from the left: $$\langle x| \hat a|\psi \rangle = \frac 1{\sqrt{2}}(x \psi(x)+\frac d{dx}\psi(x))$$ which is what was used in the book as you've mentioned yourself. Also note that $$\hat a \langle x | 0 \rangle$$ doesn't make sense, because $$\langle x | 0 \rangle \in \mathbb C$$ is not a member of $$\mathcal H$$, meaning that $$\hat a: \mathcal H \rightarrow \mathcal H$$ can't even act on it.

Check out this similar question and its answers for more details.