# Does it make sense to add eigenstate bras, such as $\langle x|+\langle p|$?

For a given state vector $$|\mathcal{S}(t)\rangle$$ in a Hilbert space, it's known that we can express it in different bases. For instance, we can express it in the position basis as $$\langle x|\mathcal{S}(t)\rangle=\Psi(x,t)$$, or the momentum basis as $$\langle p|\mathcal{S}(t)\rangle=\Phi(p,t)$$.

The bras, $$\langle x|$$ and $$\langle p|$$, are members of the dual space to the Hilbert space. Since the dual space is itself a vector space, it's closed under linear combinations, and we may write a bra such as $$\langle x+p|\equiv\langle x|+\langle p|$$. We may imagine that this, acting on some wavefunction $$\Psi$$, gives the inner product:

$$\langle x+p|\Psi\rangle=\int\left(\delta(x'-x)+\frac{1}{\sqrt{2\pi\hbar}}e^{-ipx'/\hbar}\right)\Psi(x',t)\mathrm{d}x'$$

But $$\Psi$$ has dimensions of $$\text{m}^{-1/2}$$; $$\Phi$$ has dimensions of $$(\text{kg}\cdot\text{m}\cdot\text{s}^{-1})^{-1/2}$$. So, applying the composite bra to the state vector puts us in trouble:

$$\langle x+p|\mathcal{S}(t)\rangle=\langle x|\mathcal{S}(t)\rangle+\langle p|\mathcal{S}(t)\rangle=\Psi(x,t)+\Phi(p,t)$$

This is nonsense, since the two terms in the sum have different dimension. So, what went wrong here?

• It looks like you have already shown exactly what goes wrong here. Feb 10, 2022 at 5:26
• @BioPhysicist I think I've spotted something - the momentum eigenfunction, in position basis, doesn't have dimensions of a typical wavefunction in position basis. Why is this? Feb 10, 2022 at 5:33
• There is absolutely nothing wrong with the expression $\langle x \vert + \langle p \vert$. It is a perfectly valid state vector, though this doesn't answer your units question. But I want to point out that the notation $\langle x + p \vert$ is confusing at best and ill-defined at worst; it should be avoided. Feb 10, 2022 at 5:34

First a comment on notation. It seems that you are confusing what it means to add vectors. Remember that the $$x$$ you have inside a bra or a ket is just a label, not a real number to be added to other real numbers. What it means to have a position eigenvector $$|x\rangle$$ is that

$$X|x\rangle=x|x\rangle.$$

So by the definition of addition you have presented we should have

$$|x_1\rangle+|x_2\rangle=|x_1+x_2\rangle,$$

But this is very confusing. Why would being in a superposition between $$x_1$$ and $$x_2$$ have anything to do with the completely different and uncorrelated point $$x_1+x_2$$?

When you start doing this when mixing diferent basis (i.e. position and momentum) you're bound to have a bad time.

Now for more pressing matters. There's nothing stopping you from being in a superposition between position states and momentum states, or to define vectors that are a superposition of those. You just have to be careful with normalizations. Let's say we want a vector that represents a superposition of a state $$|\Psi\rangle$$, with position wavefunction $$\psi_0(x)$$, and a state $$|\Phi\rangle$$, with momentum eigenfunction $$\phi_0(p)$$. We can then write

$$|\Psi\rangle=\int dx\,\psi_0(x)|x\rangle,$$ $$|\Phi\rangle=\int dp\,\phi_0(p)|p\rangle.$$

Here we have that these are honest-to-goodness physical states if

$$\int dx\,|\psi_0(x)|^2=\int dp\,|\phi_0(p)|^2=1.$$

Now let's look at what happens when we act on your state $$|\mathcal S(t)\rangle$$ with the bra $$\langle\Psi|+\langle\Phi|$$, the superposition of the vectors we've just defined. Linearity and the distributive property hold just fine,

$$(\langle\Psi|+\langle\Phi|)|\mathcal S(t)\rangle=\langle\Psi|\mathcal S(t)\rangle+\langle\Phi|\mathcal S(t)\rangle$$ $$(\langle\Psi|+\langle\Phi|)|\mathcal S(t)\rangle=\int dx\,\psi^*_0(x)\langle x|\mathcal S(t)\rangle+\int dp\,\phi^*_0(p)\langle p|\mathcal S(t)\rangle$$ $$(\langle\Psi|+\langle\Phi|)|\mathcal S(t)\rangle=\int dx\,\psi^*_0(x)\Psi(t,x)+\int dp\,\phi^*_0(p)\Phi(t,p),$$

which makes perfect sense and you can check that the dimensions are correct. The main issue with the dimensions is that, because $$x$$ and $$p$$ are continuous variables, they have no true eigenstates. State vectors should be dimensionless, and that does not hold for things of the type $$|x\rangle$$ and $$|p\rangle$$. But if you combine them with valid wavefunctions like $$\psi_0$$ and $$\phi_0$$, they combine to dimensionless, physically meaningful vectors.

• So would it be correct to say that the "rigged vectors" $|x\rangle$ and $|p\rangle$ carry units $\text{m}^{-1/2}$ and $(\text{kg m/s})^{-1/2}$, respectively? I understand that proper vectors are unitless here. Feb 10, 2022 at 6:20
• I'm not the original poster, but yes. Keeping in mind that $\langle x | x' \rangle = \delta(x-x')$, and the delta function as units of $L^{-1}$, it must be that $| x \rangle$ has units of $L^{-1/2}$. Similarly for $|p \rangle$.
– Zack
Feb 10, 2022 at 6:31