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?

  • $\begingroup$ It looks like you have already shown exactly what goes wrong here. $\endgroup$ Feb 10, 2022 at 5:26
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    $\begingroup$ @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? $\endgroup$
    – DanDan0101
    Feb 10, 2022 at 5:33
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    $\begingroup$ 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. $\endgroup$
    – sasquires
    Feb 10, 2022 at 5:34

1 Answer 1


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


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


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.

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    $\begingroup$ 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. $\endgroup$ Feb 10, 2022 at 6:20
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    $\begingroup$ 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 $|p \rangle$. $\endgroup$
    – Zack
    Feb 10, 2022 at 6:31

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