# QM: what is <x|p_y> = ?

1. Sep 27, 2008

### varunag

In our Quantum mechanics course we were told that the wave function can be projected in the momentum space by taking the fourier transform of the projection of the wave function in the coordinate space. This was shown in one-dimension as:
$$\langle p|\psi \rangle = \int_{-\infty}^{\infty} \langle p|x \rangle \langle x|\psi \rangle dx$$

But, how do we write if we hve to write in three dimensions?
i.e.
$$\langle \vec{p}|\psi \rangle = \int_{-\infty}^{\infty} \langle \vec{p}|\vec{r} \rangle \langle \vec{r}|\psi \rangle dx$$
Here I'm confused as to what is $$\langle \vec{p}|\vec{r} \rangle$$ ?
If I write $$\vec{p} = p_x \hat{i} + p_y \hat{j} + p_z \hat{k}$$
and $$\vec{r} = x \hat{i} + y \hat{j} + z \hat{k}$$
then should the terms like, $$\langle p_y|x \rangle$$ crop up in the final result?

Last edited: Sep 27, 2008
2. Sep 27, 2008

### maverick280857

In 3 dimensions, the position operator $\hat{x}$ is conventionally denoted by $\hat{r}$ (since the position vector in 3-space is denoted by $\vec{r}$). Having said that you can still use $\hat{x}$ with the understanding that $\hat{x}$ has three components and that the integral you have written down has to be carried out over each of the 3 spatial coordinates.

Using $\hat{r}$, I will write this as

$$\langle p|\psi\rangle = \int d^{3}\vec{r} \langle p|r\rangle \langle r|\psi\rangle$$

Now, I can use

$$\langle r|p \rangle = \frac{1}{(2\pi \hbar)^{3/2}}\exp\left(\frac{i}{\hbar}{\vec{p}\bullet \vec{r}}\right)$$

But I think what you wanted to ask was the significance of the quantity $\langle x|p_{y}\rangle$.

EDIT: To do it rigorously, you can use linearity of the inner product:

$$\langle \alpha_{1} + \alpha_{2} | \beta_{1} \rangle = \langle \alpha_{1} | \beta_{1} \rangle + \langle \alpha_{2} | \beta_{1}\rangle$$

with suitable number of terms, using the expansion of the position and momentum operators in terms of their x, y, z components and then using the expressions for $\langle x|p_{x}\rangle$, and so on. You will get the cross term $\langle x|p_{y}\rangle$ in such an expression.

Last edited: Sep 27, 2008
3. Sep 27, 2008

### varunag

The result you have given is known, but I don't know how the result came. i.e. how do we write:
$$\langle \vec{p}|\vec{r} \rangle = \langle p_x|x \rangle \langle p_y|y \rangle \langle p_z|z \rangle$$?

4. Sep 27, 2008

### maverick280857

This follows from the generalization of the Fourier Transform in 3 dimensions. Specifically,

$$\psi(\vec{r}) = \frac{1}{(2\pi\hbar)^{3/2}} \int d^{3}\vec{p} \exp\left(-\frac{i}{\hbar}\vec{p}\bullet\vec{r}\right)\phi(\vec{p})$$

$$\phi(\vec{p}) = \frac{1}{(2\pi\hbar)^{3/2}} \int d^{3}\vec{r} \exp\left(\frac{i}{\hbar}\vec{p}\bullet\vec{r}\right)\psi(\vec{r})$$

You can write this in Dirac notation easily.

Note: The result is consistent with the fact that $[x_{l},p_{m}] = i\hbar \delta_{l,m}$.

EDIT: This can be derived by inserting complete sets of states, i.e. writing $\langle \vec{p}|\vec{r}\rangle$ as an integral (multiple integral) over basis states corresponding to the different x, y, z components of p and r.