# Is this explanation of the bra-ket notation correct?

I would be very grateful for feedback, particularly pointing any factual mistakes, in the below explanation of the bra-ket notation:

"Quantum mechanical expressions can be simplified using a bra-ket notation. A basic bra-ket expression is presented below: $$I = \int f^{*}_{b} \Omega f_{k}^{} d\tau = \langle b | \Omega | k \rangle$$ The ket $$|k\rangle$$ denotes a quantum state of the system, described by the function $$f_{k}^{}$$. The bra $$\langle b |$$ denotes the state of the system that can be observed in an experiment, and it is described by the complex conjugate of the function $$f_{b}^{}$$, $$f_{b}^{*}$$. Each ket has a corresponding bra, related through the conjugate transpose, hence $$f_{k}^{}$$ = $$f_{b}^{}$$ (it should be noted here that the kets and bras can be represented as functions or vectors, since in the quantum mechanical context functions can be regarded as complex vectors of infinite dimensions). Bra is achieved by applying an appropriate operator $$\Omega$$ to the ket. An operator is a quantum mechanical equivalent of a classical function, and represents an observable in a dynamical variable, e.g. position varying with time. For example, if the initial state of the system is $$|k\rangle$$, then we can apply the position operator $$\Omega$$ to obtain a new state $$|b\rangle = \Omega | k \rangle$$, which represents the position of the particle. We can then compare the state $$|b\rangle$$ obtained from the calculation with the actual result of the experiment to see if they agree."

I am completely new to the topic of qunatum mechanics. This was gathered using mostly Atkins' Molecular Quantum Mechanics 4th edition and the Internet. I have probably checked every thread on here concerning the bra-ket notation, but the number of different terms and approaches (e.g. vectors vs functions) confuses me so much.

• Thinking of bra-operator-ket as just shorthand for an integral is useful for getting started with this notation, but it isn’t really correct. For example, when the bras and kets are representing spin states, there is no function of position corresponding to them and no integral to be performed. Dec 10, 2022 at 7:32
• @Ghoster Well, then one can generalize this notation by meaning sum or integral or both... Dec 10, 2022 at 7:33
• I think it’s best to conceptualize this in all cases as a scalar product in a complex vector space. When the vector space happens to have infinite dimension, you get an integral. Dec 10, 2022 at 7:39
• @Ghoster thank you!
– Luna
Dec 10, 2022 at 7:56
• It's not incorrect to think of bra-operator-ket as an integral, it just depends on the context. A state containing only information on spin (from which you can get the probability that you find an electron in spin up, for example) will be a vector, since there are only 2 spin outcomes. A state containing only information on position will be a function, since there are infinitely many possible positions, and we need a probability density for each one. Dec 10, 2022 at 8:24

## 1 Answer

The bra $$\langle b |$$ denotes the state of the system that can be observed in an experiment,

It is true that usually how inner products $$\langle b | \psi \rangle$$ show up is in a context where the leftmost ket is the observable quantity and $$\psi$$ is the state of the system. But just like any math you can (and probably will) see inner products calculated in which this need not be the case. The important thing to remember here is the following, which is one of the core "postulates" of quantum mechanics: If the particle is in a state $$|\psi \rangle$$, then the probability of getting some experimental outcome, say an energy of $$E$$, is equal to $$|\langle E | \psi \rangle|^2$$, where $$|E\rangle$$ is a vector corresponding to that experimental outcome. You can calculate this for specific energies, so the probability of getting outcome $$E=0.2$$ Joules is $$|\langle 0.2 | \psi \rangle|^2$$, as long as that is one possible energy that the system is compatible with. If there is a continuum of outcomes rather than a countable set of outcomes, then this gives a probability density rather than a probability. That is the reason that you have experimental outcomes showing up in inner products.

Notice, by the way, that since $$\langle f | g \rangle = \overline{\langle g|f \rangle}^*$$ for any vectors or functions $$f,g$$, then $$|\langle E | \psi \rangle|^2=|\langle \psi | E \rangle|^2$$, so in these probabilities we need not have the "experimental outcome ket" on the left, though we normally do.

and it is described by the complex conjugate of the function $$f_{b}^{}$$, $$f_{b}^{*}$$.

Given some ket $$|\psi\rangle$$, then there is a corresponding bra which we write as $$\langle \psi |$$ which is basically just the same vector, but in bra form instead of ket form. In that case the functions (which give the components of that vector) are just complex conjugates of each other. Certainly you could take the inner product $$\langle \psi | \psi \rangle$$. However in general if you have an inner product $$\langle \phi | \psi \rangle$$, the two functions can be completely different. Most physicists think of this in terms of finite-dimensional vectors, wherein if a ket is

$$|\psi \rangle = \begin{bmatrix}\psi_1 \\ \psi_2\end{bmatrix}$$

Then its corresponding bra is

$$\langle \psi | = \begin{bmatrix}\psi_1^* & \psi_2^*\end{bmatrix}$$

Bra is achieved by applying an appropriate operator $$\Omega$$ to the ket.

This is not true. Generally the bra corresponding to a ket is achieved by applying a conjugate transpose, as above.

An operator is a quantum mechanical equivalent of a classical function, and represents an observable in a dynamical variable, e.g. position varying with time.

In quantum mechanics, there is a correspondence between classical variables (not functions) and operators. Position $$x$$ has its own operator, momentum $$p$$ has its own operator, so do energy $$E$$ and angular momentum $$L$$ have their own operators. The operators don't vary in time, at least not in your introductory course which will use the "Schrödinger Picture" of quantum mechanics. Later you can change pictures, but this is not needed now.

For example, if the initial state of the system is $$|k\rangle$$, then we can apply the position operator $$\Omega$$ to obtain a new state $$|b\rangle = \Omega | k \rangle$$, which represents the position of the particle.

All correct except for the last part - the new state does not represent the position of the particle. The importance of an observable operator like the position operator $$\hat{x}$$ is basically just that we can use it to find position eigenstates $$|x\rangle$$ through the eigenvalue equation:

$$\hat{x}|x\rangle = x |x \rangle$$

The labels on that equation can be confusing because there are so many $$x$$, I'm sorry. One is an operator, one is a vector, one is just a number. Secondly, the position operator can be useful to find expected values. The average value of the position when the particle is in a state $$|\psi \rangle$$ is equal to $$\langle \psi | x | \psi \rangle$$.

We can then compare the state $$|b\rangle$$ obtained from the calculation with the actual result of the experiment to see if they agree.

No this is not correct. The comparison to experiment comes from the predictions of probabilities, which come from repeating the same experiment with the same initial conditions over and over. How to get those probabilities I answered at the beginning of this answer.

Anything I didn't comment on was correct.

• Thank you very much, your answer is extremely clear and helpful!
– Luna
Dec 10, 2022 at 10:18
• @Luna Glad to hear. If you have followup questions feel free to ask. Dec 10, 2022 at 10:49
• In fact, if you don't mind, there is something I still don't understand. I don't see why when using vectors, it seems that bra is always a conjugate transpose of ket, but when using functions, various sources (including your answer) state that the functions can be different: if $|ket\rangle = f_k$ and $\langle bra| = f_b^{*}$, $f_b$ might not equal $f_k$. I would have supposed that $f_k$ always equals $f_b$. Am I misunderstanding something?
– Luna
Dec 10, 2022 at 11:17
• It depends which bra and ket. For each (vector or function) as a ket $|\psi \rangle$ , there is a special bra which is the same (vector or function) but in bra form, and its components are the complex conjugates of the original components. In some contexts that special bra is called the dual of that ket. In that case you have $f_k = f_b$ by your definition of those functions in the comment above. However just like there are many kets (which we may represent by column vectors), there are just as many bras (which we may represent by row vectors). Dec 10, 2022 at 11:24
• Of course, thank you!!
– Luna
Dec 10, 2022 at 11:30