I'm confused about bra-ket reversal, and why it requires conjugation.

$$ \langle \phi|\chi \rangle = \langle \chi|\phi \rangle ^* $$

Or equivalently why

$$ |\chi \rangle ^* = \langle \chi | $$

I tried to understand it with the following argument (which follows Feynman Lecture's Vol 3 Ch 5). If you have the Feynman lectures, I'm trying to understand how Feynman gets equation 5.26.

Lets say $R$ is a particular complete three-dimensional basis for the states our system can take (like for a spin-1 particle), with basis vectors $R_1,R_2,R_3$.

A state $\chi$ can be represented in this basis since the basis is complete: $$ |\chi\rangle=\sum_{i=1,2,3}|R_i\rangle\langle R_i | \chi \rangle $$

Now consider the quantity $$\langle \chi | \chi \rangle$$ Physically, we would say this is the complex amplitude for a system in state $\chi$ to be in state $\chi$. It is just a number (which could be complex if we haven't assumed $|\chi \rangle ^* = \langle \chi |$).

Since the probability associated with this amplitude is the norm (length) of the amplitude, we can say confidently that whatever number $\langle \chi | \chi \rangle$ is, its norm is one: $$ \langle \chi | \chi \rangle^*\langle \chi | \chi \rangle=1 $$ In words, there is 100% probability the system is in the state it is in.

But it seems to me that only restricts it to the unit circle in the complex plane: $\langle \chi | \chi \rangle = e^{i\theta}$ where $\theta$ could be anything.

If we write $\langle \chi | \chi \rangle$ in our $R$ basis, we get $$ \langle \chi | \chi\rangle=\sum_{i=1,2,3}\langle \chi |R_i\rangle\langle R_i | \chi \rangle = e^{i\theta} $$ We see that each term $\langle \chi |R_i\rangle\langle R_i | \chi \rangle$ can only vary (for different choices of $R$) in ways that preserve the last equality, but it seems like requiring

$$ \langle R_i|\chi \rangle = \langle \chi|R_i \rangle ^* $$

Is too strong a condition if $\theta$ is arbitrary. If $\theta=0$ or you needed to pick a value and stick with it, I could see why that condition may be required, but I'm not even sure about that. Or maybe I just don't see the argument.

Can anyone help with the reasoning here?

In Feynman lectures, he says effectively that because the probability for $\chi$ to be any state in $R$ is 1 (R is a complete basis)

$$ \langle R_1 | \chi \rangle \langle R_1 | \chi \rangle ^ * + \langle R_2 | \chi \rangle \langle R_2 | \chi \rangle ^ * + \langle R_3 | \chi \rangle \langle R_3 | \chi \rangle ^ * = 1 $$ and because the amplitude $\langle \chi | \chi \rangle = 1$... $$ \langle \chi | R_1 \rangle \langle R_1 | \chi \rangle + \langle \chi | R_2 \rangle \langle R_2 | \chi \rangle + \langle \chi | R_3 \rangle \langle R_3 | \chi \rangle = 1 $$

Then because $R$ and $\chi$ can freely vary, I guess (?) each term could maybe vary independently requiring for each term that $$ \langle \chi | R_i \rangle = \langle R_i | \chi \rangle ^* $$

If this were not true, probability wouldn't be "conserved," and particles would get "lost." -Feynman

but this does assume the amplitude in the second equation is $1$ (not $e^{i \theta} $ ) while it safely assumes the probability in the first is $1$.

Care to help me understand this argument?

Someone could instead prove that any choice other than $1$ would result in absurd physics. Or someone could explain that $\langle \phi|\chi \rangle = \langle \chi|\phi \rangle ^*$ results from a deeper symmetry.

  • $\begingroup$ What would happen to probability if $\langle x|x \rangle $ was not equal to 1? $\endgroup$
    – user108787
    Aug 29, 2016 at 1:47
  • $\begingroup$ That's a good question. Seems to me like probability can only continue working if you pick a value of $e^{i \theta} $ instead. $\endgroup$
    – JoseOrtiz3
    Aug 29, 2016 at 1:49

3 Answers 3


Recall the definition of an inner product space, a vector space equipped with a map $n$ that returns a scalar for any pair of vectors. In quantum mechanics, the Hilbert space of states for any quantum system is always an inner product space with the interpretation you noted above: The inner product of a state $A$ on a state $B$, $\langle A|B\rangle$, is the probability amplitude of measuring the system to be in state $A$ given you have knowledge the system is in state $B$.

Rather than considering the set of vectors and a two vector map $n(w,v)$, we can take the perspective instead that there is a dual space. Physical state vectors are "kets" if you will, but there exists another vector space, called the dual space, which is filled with "bras". These bras form the space of linear maps from the kets to complex numbers (the scalars in quantum mechanics).

Notice that there is a one-to-one correspondence between the bras and kets since, by linearity, operators in the dual space are fully defined by their action on a complete set of basis vectors in the ket space. That means that we can have a notion of taking a ket to a bra and back, normally we denote that by $\dagger$, but also sometimes by $*$ or a bar.

And, again using the correspondence, we parameterize the space of bras by objects, $\langle x |$, with the property $\langle x | y \rangle = \delta_{xy}$. Immediately we can work out the transformation properties for an orthonormal basis of bras by considering the action of a properly normalized bra on a properly normalized ket

$$|x\rangle = a_1 |a_1\rangle + a_2 |a_2\rangle.$$

The dual vector must be given by

$$\langle x | = \bar{a}_1 \langle a_1 | + \bar{a}_2 \langle a_2| $$

if we are to have that

$$\begin{align}\langle x|x\rangle &= \bar{a_1} a_1 \langle a_1|a_1\rangle + \bar{a_2} a_2 \langle a_2|a_2\rangle + \bar{a_1} a_2 \langle a_1|a_2\rangle + \bar{a_2} a_1 \langle a_2|a_1\rangle \\ &= \bar{a_1} a_1 \langle a_1|a_1\rangle + \bar{a_2} a_2 \langle a_2|a_2\rangle + 0 + 0 \\ &= 1,\end{align}$$

which is required by orthonormality.

  • $\begingroup$ Thank you. Two questions: "$\langle A | B \rangle$ is the probability of measuring the system to be in state...." Don't you mean amplitude, a complex number? You still have to take the norm of that to get the probability, right? Also, it seems we are still assuming the amplitude $\langle x | x \rangle = 1$ and not $e^{i \theta}$, both of which have norms of $1$. Also I might need clarification on "into the underlying field" $\endgroup$
    – JoseOrtiz3
    Aug 29, 2016 at 2:25
  • $\begingroup$ Yes, $\langle A|B\rangle$ is of course the amplitude. Underlying field means the scalars of the vector space. You are right that we are choosing a basis for our dual space such that the operator's act on our vectors in an orthonormal way. In general, there is no absolute necessity for this. However, if we chose to use a basis where $\langle x|x\rangle = k$, then we simply make computations more difficult. In general, we could use a basis such that $\langle x|x\rangle = k(x)$, but then we would only complicate our lives since there seems to be no physics in the choice of $k$, so we choose $k=1$ $\endgroup$ Aug 29, 2016 at 18:40

But it seems to me that only restricts it to the unit circle in the complex plane: ⟨χ|χ⟩=eiθ where θ could be anything.

The probability of finding a system, prepared in state $|y\rangle$, to be in $|x\rangle$ is

$$P(y \rightarrow x) = \frac{|\langle x | y \rangle |^2}{\langle x | x \rangle \langle y | y \rangle }$$

The probability must be real and the numerator is real. It follows that $\langle x | x \rangle$ and $\langle y | y \rangle$ are either real or complex conjugates.

But then

$$P(y \rightarrow y) = \frac{|\langle y | y \rangle |^2}{\langle y | y \rangle \langle y | y \rangle } = 1$$

which implies $\langle y | y \rangle$ is real.

  • $\begingroup$ $P(y \rightarrow x) = \frac{|\langle x | y \rangle |^2}{\langle x | x \rangle \langle y | y \rangle }$ Where did the denominator come from? Why to find the probability are you dividing by two amplitudes? I have only ever seen $P(y \rightarrow x) = |\langle x | y \rangle |^2$ which I guess assumes the vectors are normalized. $\endgroup$
    – JoseOrtiz3
    Aug 29, 2016 at 2:55
  • $\begingroup$ @OrangeSherbet, see, for example, this screen shot from Weinberg's lectures on QM $\endgroup$ Aug 29, 2016 at 3:34
  • $\begingroup$ Interesting. I suppose my next question is where does such a formula become useful? If $\langle x | x \rangle$ is always equal to $1$, then the formula is only different from the simpler $P(y \rightarrow x) = |\langle x | y \rangle |^2$ in cases where it's not 1. I guess I just don't see reason to have that product in the denominator instead of just assuming its 1. $\endgroup$
    – JoseOrtiz3
    Aug 29, 2016 at 17:37
  • $\begingroup$ @OrangeSherbet You're right, if vectors are normalized then $\frac{1}{\sqrt{\langle y | y \rangle}} |y\rangle = |y\rangle$, which gives the version you stated. Mr. Alfred's notation generalizes to the case where they aren't normalized. $\endgroup$ Sep 6, 2021 at 6:09

Sorry, my tablet was charging earlier and the stylus was all over the place.

You have pretty much the answer by bobak that I would have written. So say you have the simplest normalised function $Ae^{i\theta }$. The conjugate $Ae^{-i\theta }$ makes sure that you are left with $A^2$ and not .5 A or 6A or anything that would screw up the probabilities adding up to 1, when later you need to expand the orginal function into its base kets, if you follow me, to obtain the probability that the system is in a given eigenstate.

It's just the complex conjugate version of the scaler/ dot product. Say you have a vector $B = 5e_x + 6e_y+7e_z $ and you want to extract the component of $e_y $, so you do the dot product of $B.e_y $ and out pops 6.

So replace the dot product with $\langle x | x \rangle $ and the component 6 by the amplitude (whatever it is).

This thing $\langle x | x \rangle $ gets rid of the other bases since $\langle x | y \rangle $ = 0 and keeps the amplitude untouched by making sure $\langle x | x \rangle $ = 1. I am mixing functions with variables in the kets but i reckon you should have the general idea when you practice a bit.

I hope this makes sense

  • $\begingroup$ I like the idea of "leaving the amplitude untouched by making sure $\langle x|x \rangle = 1$", it sounds like if it were instead a complex number, then phases between amplitudes for different paths could not be compared with each other, since a path where nothing occurred may arbitrarily interfere with another path where nothing occurred twice, for instance. Maybe this is getting somewhere? $\endgroup$
    – JoseOrtiz3
    Aug 29, 2016 at 3:02

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