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Sorry if this has been asked before in some manner, but I'm just a bit confused about the distinction between a state $\alpha$ and its ket $|\alpha\rangle$. I was recently told that a state $\alpha$ is an element of the Hilbert space $\mathcal{H}$, and a ket can be interpreted as a linear map $\mathbb{C} \rightarrow \mathcal{H}$ with action $c |\alpha\rangle = c \alpha$ for $c \in \mathbb{C}$. What then, is the difference between the two objects $\alpha$ and $|\alpha\rangle$?

Say if you had some decomposed operator $O$ acting on an element of the basis of $\mathcal{H}$ given by $\{e_i\}$ (and assume $\psi = e_j$, and the $\{p_i\}$ are elements of $\mathbb{C}$): $$ O \psi= \sum_i p_i |e_i\rangle \langle{e_i}| e_j = \sum_i p_i |e_i\rangle \delta^i_j = p_j |e_j\rangle $$ Is this new object $p_j |e_j\rangle$ meaningfully different from the state $p_j e_j$, since technically we have defined a ket as a linear map and not as an element of the state space? If it is, then wouldn't that mean $O$ would not have its action as $\mathcal{H} \rightarrow \mathcal{H}$? If they are the same, how does the adjoint fit into this definition, i.e. if a bra is the adjoint of its corresponding ket, is there some meaning to taking the adjoint of a state?

Apologies again if this has been asked before, thanks in advance.

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    $\begingroup$ Related: Hilbert space vs. Projective Hilbert space “$c\lvert \psi \rangle$ is the same state as $\lvert \psi \rangle$” $\endgroup$
    – Ghoster
    Mar 3 at 1:41
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    $\begingroup$ I assume that whoever gave you the definition of ket told you the meaning of $c |\alpha\rangle = c \alpha$ . I have a hard time understanding it. $\endgroup$ Mar 3 at 4:14
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    $\begingroup$ Kets are vectors in a Hilbert space. States are equivalence classes of kets related by scalar multiplication by complex numbers. $\endgroup$
    – Ghoster
    Mar 3 at 4:23
  • $\begingroup$ @GiorgioP-DoomsdayClockIsAt-90 It actually is a common definition if properly phrased, perhaps in a more mathematical oriented texts. I.e., for any $\psi\in H$, you can define $|\psi\rangle: \mathbb C \to H$ by $|\psi\rangle(\lambda):=\lambda \psi$. And with this definition, it indeed can be shown that the bra, i.e. the linear functional associated to $\psi$ is the adjoint of $|\psi\rangle$. :) $\endgroup$ Mar 3 at 8:32
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    $\begingroup$ @Ghoster See my comment above. It depends on how you define things (as always), so there is no universal answer here. And OP should not get confused by different definitions, which I guess is the problem here. The point is that the relation between $\psi\in H$ and $|\psi\rangle \in L(\mathbb C, H)$ is via an isomorphism, so both objects "contain the same information" and it does not matter too much with what object we work with. $\endgroup$ Mar 3 at 8:35

5 Answers 5

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I think all other answers so far do not address the problem directly. The definition of the OP regarding kets is a quite common, though not the most used one, see e.g. references 1 and 2. So I don't think one should argue now which definition of "ket" is the right one (the question is ill-posed: it is just notation).

The OP further uses the word "state" to refer to an element of the Hilbert space, which a priori is fine, but it does not correspond to the physical notion of the word (but to be fair, I think many intro texts use this notion, although one might add a normalization constraint). Others have commented on that, although it is not the primary question and actually not relevant here.

The question asks about the difference between $\psi\in H$ and $|\psi\rangle\in L(\mathbb C,H)$, based on a confusion and mixing of different notations and concepts.


To start, let me recap the basics which lead to the notation introduced in the question: For any vector $v\in H$ of a complex Hilbert space $H$, we can define a map $|v\rangle: \mathbb C \to H$ by $|v\rangle: \lambda \mapsto \lambda v$. In fact, the thus defined map: $\mathcal I: H\to L(\mathbb C, H)$, with $\mathcal I: v\mapsto |v\rangle$ is a canonical isomorphism. This means, roughly speaking, that we can relate both objects $v$ and $|v\rangle$ in a natural way, without "loosing information". Further, with this definition, we can meaningfully speak of the adjoint of $|v\rangle$, which turns out to be the well-known "bra", denoted by $\langle v|$. As a last point, let me remark that it is exactly this notion of a "ket", which people refer to when saying things like "the bra is the adjoint of the ket" or so.


Let me now clear up the confusion in the question:

With the definition of "ket" given in the question, we have to make sure to understand all other symbols accordingly. Most importantly, if we write $|\psi\rangle\langle\psi|$, then we denote by this the composition of the "ket" and the "bra" map, i.e. $|\psi\rangle\langle\psi|:=|\psi\rangle \circ \langle \psi|: H \to H$, with $$|\psi\rangle\langle\psi| (v)= |\psi\rangle\left(\langle\psi,v\rangle_H\right)= \langle\psi,v\rangle_H \,\psi \quad . \tag 1$$

Hence, in this notation we have that $|\psi\rangle\langle\psi|$ acts as a hermitian one-dimensional projection on $\psi \in H$, which often is also written as $P_\psi$ (which makes no reference to any ket-notation). The spectral theorem now allows to decompose every hermitian operator $A$ as

$$ A = \sum\limits_{j=1}^{\dim H} a_j\, P_{\psi_j}\overset{(1)}{=}\sum\limits_{j=1}^{\dim H} a_j\, |\psi_j\rangle\langle\psi_j| \quad ,\tag 2$$

where the $\psi_j$ are (orthonormal) eigenvectors of $A$ with eigenvalues $a_j$. The first equality is the spectral theorem, the second a consequence of our notation and definitions.

We therefore obtain

$$A v\overset{(2)}{=}\sum\limits_{j=1}^{\dim H} a_j\,|\psi_j\rangle\langle\psi_j| (v)\overset{(1)}{=}\sum\limits_{j=1}^{\dim H} a_j\, \langle \psi_j,v\rangle_H\, \psi_j \tag 3 \quad ,$$

and everything makes sense, i.e. the linear operator $A$ maps vectors of the Hilbert space $H$ to vectors again.


Now, the reason many people seem to be confused here is, I suppose, that in most physics books a "ket" is defined to be a vector in the Hilbert space. Put differently, then it is assumed that $|\psi\rangle \in H$ and symbols like $\psi$ have no meaning; it is the whole object, the ket, which is the vector. Then the operator $P_\psi$ can also be written with this notation as $P_\psi=|\psi\rangle\langle \psi|$, where the symbol on the RHS is defined as the operator $|\psi\rangle\langle \psi|: H\to H $ with

$$ |\psi\rangle\langle \psi|: |v\rangle \mapsto \langle \psi|v\rangle |\psi\rangle \quad , \tag 4$$

just as before (if you identify both ways to write the inner product); but to emphasize: The symbols here have a different meaning than before. It is thus important to not mix both notations and concepts, and stick to one convention from the beginning. Else one might end up in contradictions and ill-defined expression as in the question.

You do not get or lose anything by using one or the other notation, a priori. For some purposes one might be more suitable than the other. On the other hand, many people (including myself) do not use any of these "bra-ket" notations at all.


References:

  1. Quantum Information Theory. J. M. Renes. De Gruyter. Appendix B2, p. 276.
  2. Categories for Quantum Theory: An Introduction. C. Heunen and J. Vicary. Section 0.2.4, p. 18.
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    $\begingroup$ In (1), what is the difference between $|\psi\rangle$ and $\psi$? That is what the OP asked, and you don’t seem to have explained it. $\endgroup$
    – Ghoster
    Mar 3 at 18:49
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    $\begingroup$ The question asks about states, so not commenting on states doesn’t answer the question. $\endgroup$
    – Ghoster
    Mar 3 at 18:52
  • $\begingroup$ @Ghoster No, I think you misunderstand/I've written it poorly. The OP defines "state" to be a vector in a Hilbert space--which is fine, but it is non-standard in the physical sense of the word state. Other answers have pointed that out --but this was not the question of the OP. The OP got confused because they mixed different "ket"-notations, namely both I discuss, and arrived at some contradiction, namely: operator maps vector to linear map from complex numbers to Hilbert space. I have tried to edit my post. $\endgroup$ Mar 3 at 19:08
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    $\begingroup$ @AlbertusMagnus Because I don't like it, and in my experience it only leads to confusion, such as thinking that the adjoint operator acts on the bras or similar things, without adding too many useful things. I also think that many people fail to see that Dirac notation, a priori, is only this: a notation. But again, this is only my personal experience/ opinion. I don't use any particular notation, just the usual math notation in linear algebra or functional analysis. $\endgroup$ Mar 4 at 8:12
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    $\begingroup$ @GiorgioP-DoomsdayClockIsAt-90 Let me add: My answer here should in no way be an "advertisement" for this notation; rather, I think all answers before just ignored the fact that OP uses this notation and then got confused because of an inconsistent usage. $\endgroup$ Mar 4 at 8:46
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This is really two questions in one, namely what is a ket and what is a state. Both questions are bound to cause some degree of controversy between physicists of different mathematical pedigrees. I will assume finite dimensional Hilbert spaces for the entirety of this answer, since that contains the relevant conceptual details - infinite dimensions are morally the same, but come with significantly increased technical baggage.


Kets

Def: A ket is an element of the Hilbert space $\mathscr H$, and a bra is an element of the dual space $\mathscr H'$.

This definition is straightforward, and is basically defining a convenient notation. From this perspective, $|a\rangle$ is simply a vector in $\mathscr H$, and $a$ is a label we have chosen for it. This is particularly convenient when talking about eigenvectors; if a particular vector is an eigenvector of some operator $A$ with eigenvalue $a$, it is convenient to call that vector $|a\rangle$ as a helpful reminder of its most contextually relevant characteristic (namely, that $A|a\rangle = a |a\rangle$). Note that $a$ is a number, while $|a\rangle$ is the ket (i.e. element of $\mathscr H$) which we have labeled by that number.

Let's say $\psi_a$ and $\psi_b$ are eigenvectors of some self-adjoint operator $A$ with eigenvalues $a$ and $b$, respectively. We will write them in "ket form" as $|a\rangle$ and $|b\rangle$. The action of the bra $\langle a|$ on the ket $|b\rangle$ is given by

$$\langle a| \big( |b\rangle\big) \equiv \langle a|b\rangle = \langle \psi_a,\psi_b\rangle$$

Therefore, we can compute the inner product between $|a\rangle$ and $|b\rangle$ by turning $|a\rangle$ into its bra "dual" $\langle a|$ and then operating on $|b\rangle$.

This notation is exceedingly convenient. As an example (which we will see again in a bit), we might consider the operator which projects an arbitrary vector onto (the linear span of) $\psi_a$:

$$ \phi \mapsto \frac{\langle \psi_a, \phi\rangle}{\langle \psi_a,\psi_a\rangle }\psi_a$$

Utilizing bra-ket notation, this operator takes the straightforward form $$\frac{|a\rangle\langle a|}{\langle a|a\rangle}$$

Infinite dimensional caveat: One of the subtleties which arises in infinite dimensions is the fact that we have to consider generalized (or non-normalizable) eigenvalues and eigenvectors, corresponding to operators with continuous spectra. Some authors define bras and kets such that they need not actually belong to the Hilbert space itself, as long as they obey some looser requirements. This is what happens with the "position eigenkets" $|x\rangle$, for example. For more, see e.g. my answer here.


States

This answer is more subtle, so let's zoom out a bit. In classical mechanics, a state is generally specified by giving the values of all of the possible observables which we might measure. For example:

  • For a system consisting of a single point particle we might specify a state by the values of the position $\mathbf x$ and the momentum $\mathbf p$; any other observable (kinetic energy, angular momentum, etc) is then determined by these values.

  • For a more complex system (e.g. a rigidly-rotating sphere), we might specify its center-of-mass position $\mathbf X_{cm}$, its total linear momentum $\mathbf P = M \dot{\mathbf X}_{cm}$, and its angular momentum $\mathbf L$.

When we learn statistical mechanics, we discover that it's often extremely useful to generalize this. Rather than defining a state as a specification of the precise value for each observable, we can specify a state by providing a probability distribution for the values of each observable. From this perspective, the fundamental questions we can ask take the following form:

$\text{What is the probability that the observable $\mathcal O$ takes its value in some subset $E\subseteq \mathbb R$?}\tag{$1$}$

This way of thinking is deeply physical, in the sense that this is the only thing we can ever actually measure. If we imagine that we can know the exact values of all of the relevant observables, then these probabilities are always either 0 or 1; if the exact value of $\mathcal O$ is 5, then the probability that $\mathcal O$ takes its value in $E$ is 1 if $5\in E$ and 0 otherwise. However, once we add in uncertainty, the probabilities become non-trivial.

All that is to say, a very physical definition of a state is that it is a prescription of a probability distribution to each possible observable - in other words, a mechanism for answering all possible questions of the form $(1)$.


In quantum mechanics, this definition continues to hold. Given the standard formulation of quantum mechanics in which observables are represented by self-adjoint operators on some Hilbert space $\mathscr H$, whose spectra (loosely, eigenvalues) correspond to the possible measurement outcomes, one might ask how we can assign probability distributions to observables.

Here is one possibility, in which I restrict my attention to finite-dimensional Hilbert spaces for simplicity (the fundamental idea remains the same for infinite-dimensional spaces). Any self-adjoint operator $A$ can be written in the form $A = \sum_{i} \lambda_i \mathbb P_i$, where $\lambda_i$ is the $i^{th}$ eigenvalue of $A$ and $\mathbb P_i$ is the self-adjoint projection operator onto the corresponding eigenspace. Furthermore, the sum $\sum_i \mathbb P_i = \mathbf 1$ (the identity operator).

Example:

$$A = \pmatrix{1 & 0 \\ 0 & -1} = (1) \cdot \pmatrix{1&0\\0&0} + (-1) \cdot \pmatrix{0&0\\0&1}$$

$$B = \pmatrix{0 & i \\ -i & 0} = (1)\cdot\pmatrix{1/2 & i/2 \\ -i/2 & 1/2} + (-1)\cdot \pmatrix{1/2 & -i/2 \\ i/2 & 1/2}$$

It is a straightforward exercise to demonstrate that each matrix written on the right-hand side of the above expressions is indeed a projection operator onto the appropriate eigenspace, and that the projection operators sum to the identity operator.

This structure, which underlies all self-adjoint operators, is neatly summarized in the form of a projection-valued measure $\mu$, which consists of the following map. For any subset $E\subseteq \mathbb R$,

$$ \mu(E) = \sum_{\lambda_i \in E} \mathbb P_i\tag{$2$}$$

In words, it is the sum of all of the projection operators which correspond to eigenvalues which lie in $E$.

Example:

In the case of the operator $B$ given above, we would have $$\mu_B(\{1\}) = \pmatrix{1/2&i/2\\-i/2&1/2} \qquad \mu_B\big((-\infty,0)\big) = \pmatrix{1/2&-i/2\\i/2&1/2}$$ $$\mu_B\big(\{0\}\big) = \pmatrix{0&0\\0&0} \qquad \mu_B(\mathbb R) = \pmatrix{1/2&i/2\\-i/2&1/2}+\pmatrix{1/2&-i/2\\i/2&1/2} = \pmatrix{1&0\\0&1}$$

With the concept of a projection-valued measure in hand, I make the following claim. If you pick out any non-zero vector $\psi\in \mathscr H$, the following constitutes a probability distribution for the observable represented by the self-adjoint operator $A$:

$$ \mathrm{Prob}_A(E) := \frac{\langle \psi, \mu(E) \psi\rangle}{\langle \psi, \psi\rangle}\tag{$3$}$$

Observe that $\mathrm{Prob}_A(\mathbb R) = 1$, which comes from the fact that $\mu(\mathbb R) = \mathbf 1$. It is a good exercise to prove that:

  • $\mathrm{Prob}_A(E^c) = 1-\mathrm{Prob}_A(E)$, where $E^c$ is the complement of $E$

  • $E\cap F = \emptyset \implies \mathrm{Prob}_A(E\cap F) = \mathrm{Prob}_A(E) + \mathrm{Prob}_A(F)$

So with this in mind, do vectors in the Hilbert space correspond to states? The answer is almost, but not quite - observe that for any nonzero $\alpha\in \mathbb C$, $\psi$ and $\alpha \psi$ yield exactly the same probability distribution, and therefore exactly the same physical state. As a result, a state is associated not to a single vector $\psi$ but rather the set $\{\alpha \psi \in \mathscr H : \alpha\in\mathbb C\}$. Such a set is called a ray in $\mathscr H$.

Since states correspond to entire rays, you are free to pick any vector inside the ray to use to calculate your probabilities. It is particularly (computationally) convenient if the denominator in $(2)$ is equal to 1 - i.e. if your chosen ray representative is normalized. For this reason, it is not uncommon to simply say that states correspond to normalized elements of $\mathscr H$, though this is not quite true, since even if $\psi$ is normalized, $\psi$ and $e^{i\beta}\psi$ correspond to the same state for all $\beta\in \mathbb R$.


This isn't quite the end of the story. Note that we may utilize the bra-ket notation to define the following operator:

$$\rho = \frac{|\psi\rangle\langle \psi|}{\langle\psi|\psi\rangle}$$

in which case

$$\mathrm{Prob}_A(E) = \mathrm{Tr}\big(\mu(E)\rho\big)$$ as can be straightforwardly shown in a line or two. Observe that for any $\psi,\psi'$ in the same ray, $\rho_\psi=\rho_\psi'$ - therefore, we might use this object - called a density operator or density matrix - as an unambiguous way to define a state.

However, this isn't the most general possible definition. If the state can be defined in this way (using a single ray in the Hilbert space) it is called pure. However, for any set of pure density operators $\{\rho_i\}$, the operator $$\rho = \sum_k p_i \rho_i, \quad p_i\in[0,1] \quad \sum_i p_i = 1$$

also satisfies the requirements of a state. Such a state is called mixed.

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    $\begingroup$ Excellent presentation! So, when people refer to the "adjoint of a state" they are referring to the adjoint of the density operator and not to $|\alpha\rangle^{\dagger}$? $\endgroup$ Mar 3 at 23:45
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    $\begingroup$ @AlbertusMagnus No, the density operator is self-adjoint. When people refer to the adjoint of a ket $|\alpha\rangle$, they are referring to the corresponding bra $\langle \alpha|$, despite the fact that this a bit of an abuse of terminology. The main thing to understand is that there is an unambiguous association between bras and kets which is provided by the inner product, so you can map any ket onto a corresponding bra and vice-versa. $\endgroup$
    – J. Murray
    Mar 4 at 4:05
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    $\begingroup$ @AlbertusMagnus To add to J. Murrays (correct) reply: Just to emphasize: The ket-bra notation is just notation. The correspondence between vectors of a Hilbert space and vectors of its (continuous) dual is governed by the Riesz representation theorem. $\endgroup$ Mar 4 at 8:26
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A ket is an element of a Hilbert space.

A ket may also be called a state vector.

An operator acting on a ket produces a ket.

Suppose we have a two-dimensional Hilbert space and a Hermitian operator $\hat{S}_z$. The eigenvalue equation for this operator is written $$ \hat{S}_z |\lambda\rangle = \lambda |\lambda \rangle $$ where $\lambda$ is an eigenvalue and $| \lambda \rangle$ is the eigenket (or eigenstate vector, often abbreviated to eigenstate). For example if the eigenvalues are $\pm 1/2$ we might write $$ \hat{S}_z |+\rangle = (1/2) |+\rangle, \\ \hat{S}_z |-\rangle = -(1/2) |-\rangle. $$

A general state can be written $$ | \psi \rangle = a |+ \rangle + b |- \rangle. \tag{1} $$

We can also introduce a vector notation. This is done by first adopting a basis. If the basis is $\{| +\rangle, |-\rangle \}$ then the components of the vector are $\langle + | \psi \rangle$ and $\langle - | \psi \rangle$. One may write the vector as $$ \left( \begin{array}{c} \langle + | \psi \rangle \\ \langle - | \psi \rangle \end{array} \right). $$ For the example state $| \psi\rangle$ given in equation (1) this would be $$ \left( \begin{array}{c} a \\ b \end{array} \right). $$ In this basis the ket $|+\rangle$ is represented by the vector $$ \left( \begin{array}{c} 1 \\ 0 \end{array} \right) $$ and the ket $|-\rangle$ is represented by the vector $$ \left( \begin{array}{c} 0 \\ 1 \end{array} \right). $$

In the standard use of Dirac notation one should never find oneself writing any of the following $$ | \psi \rangle = \hat{Q} \qquad \mbox{a ket cannot equal an operator} \\ | \psi \rangle = c \qquad \mbox{a ket cannot equal a number} \\ $$

It follows that one would never write $c | \alpha \rangle = c \alpha$ since the notation is not one in which such an equality makes any sense.

If one makes a distinction between 'state' and 'state vector' then, in physics at least, the distinction is ordinarily that 'state' refers to the physical system itself as a physical thing with a physical nature and 'state vector' refers to the mathematical abstraction being used to model the state.

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    $\begingroup$ I would say this corresponds nicely to what one would see in quantum physics books. A ket is a vector, which is good since there is no contradiction with understanding kets as linear functionals since they too are vectors. However, the OP is looking for a distinction between states and kets. Your post seems to make no distinction between a state and a ket? $\endgroup$ Mar 3 at 14:13
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A ket is an element of the Hilbert space, $| \psi \rangle \in \mathcal{H}$, i.e., it is an element (a vector) of the complex inner-product space. I think this is well-accepted terminology (introduced by Dirac) and to give the name "ket" to anything else would be highly non-standard. The Hilbert space represents the quantum system and the ket is sometimes called the "state" of that system. Operators on the Hilbert space represent maps from one ket to another; Hermitian operators represent observables (whose real eigenvalues are the possible measured values of that observable).

Two other definitions of a state are often made, however:

  • We can still identify the state with a ket, but recognize that the state is physically unchanged by multiplication with a complex number $a\in \mathbb{C}$. Therefore $a | \psi \rangle$ and $| \psi \rangle$ represent the same state, and we could say that the state is a ray.
  • When a system is represented by a ket it is called a pure state, sometimes expressed as a coherent superposition of other kets. A statistical (non-coherent) mixture can be represented by density matrix. For a mixture of pure states, this would take the form $\rho = \sum_i p_i \, |\psi_i \rangle \langle \psi_i|$, with $\sum p_i = 1$; but in general the density matrix could have "off-diagonal" terms, e.g., $|\psi_i\rangle \langle \psi_j|$. This operator on the Hilbert space is often called the state of the system.
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  • $\begingroup$ So, when one talks about the "adjoint of a state", they are referring to the adjoint of the density operator, and not for example, $|\alpha\rangle^{\dagger}$? To be clear, $|\alpha\rangle^{\dagger}=\langle\alpha |$ is the adjoint of the "state ket". $\endgroup$ Mar 3 at 23:39
  • $\begingroup$ Given a linear operator on a Hilbert space, $A: \mathcal{H} \rightarrow \mathcal{H}$, its Hermitian adjoint, $A^\dagger$, is the linear operator for which $\langle A^\dagger (u) , v \rangle = \langle u, A(v) \rangle$ for all $u,v \in \mathcal{H}$. In bra-ket notation, if $|\psi\rangle = A | \phi \rangle$, then $\langle \psi | \sigma \rangle = \langle \sigma | \psi \rangle^* = \langle \sigma | A(\phi) \rangle^* = \langle A^\dagger(\sigma) | \phi \rangle^* = \langle \phi | A^\dagger(\sigma) \rangle$. So $|\psi \rangle = A | \phi \rangle$ implies $\langle \psi | = \langle \phi| A^\dagger$. $\endgroup$
    – Ben H
    Mar 4 at 1:33
  • $\begingroup$ The bra associated to a ket, $| \psi \rangle$ is an element of the dual space, i.e, it is a linear map $\langle \psi| : \mathcal{H} \rightarrow \mathbb{C}$. Specifically, it is the linear map for which $\langle \psi | \left( | \phi \rangle \right) =: \langle \psi | \phi \rangle$ is equal to $\langle | \psi \rangle, | \phi \rangle \rangle$, for all $| \phi \rangle \in \mathcal{H}$. "Bras" are an inner product waiting for another vector: $\langle \psi | = \langle \psi, \cdot \rangle$ $\endgroup$
    – Ben H
    Mar 4 at 1:40
  • $\begingroup$ So, it doesn't make sense to say the adjoint of a ket/bra. They can be thought of as linear maps, but it not a linear operators on $\mathcal{H}$, they are maps from $\mathcal{H}$ (for the bra, or from the dual of $\mathcal{H}$ for the ket) to the complex numbers. One can just use the definition of the bra and the definition of the adjoint together, as I did in my first comment. $\endgroup$
    – Ben H
    Mar 4 at 1:45
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    $\begingroup$ @AlbertusMagnus See my answer. You can give the phrase "bra is adjoint of ket" a meaning, if you define ket not as a vector in the Hilbert space, but using a natural isomorphic object. The adjoint of a density operator is the density operator, since these are self-adjoint by construction. $\endgroup$ Mar 4 at 8:19
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Besides the correct introduction to the bra-ket notation done by @AndrewSteane, a cleaner mathematical notation may help to understand the situation and how manipulations go.

Kets are elements of the Hilbert space. The map you were told about is the linear map from $\mathbb{C} \times H$ to $H$ that associates to every pair $(c,|v\rangle )$ in $\mathbb{C}\times H$ the ket $c|v\rangle$ in $H$ (notice the difference of notation). Therefore, to each ket $|v\rangle$ we can associate a canonical isomorphism in $L(\mathbb{C},H)$, $\Phi(|v\rangle)$ : $c \rightarrow c|v\rangle$. Being $H$ an inner product space, the adjoint of $\Phi(|v\rangle)$ coincides with the bra $\langle v|$. More mathematical details on the MSE site may be found in this answer.

Now, let me come to your main questions.

  1. What is the distinction between a ket and a state in QM?

In a strict approach following Dirac's symbolism, a ket is an element of the Hilbert space where operators representing observables act. It is almost a synonym of state (maybe better, of a pure state). Almost, because all the normalized states differing by a phase are considered equivalent.

However, from a more general point of view, a Hilbert space of quantum mechanics can represent the $C^*$-algebra $A$ of the observables. Within the $C^*$-algebra approach, a physical state is a positive functional on $A$. In this sense, one could make a difference between states and kets.

  1. About your example of formal manipulation

The example you propose doesn't consistently use Dirac's notation. Instead of $$ O \psi= \sum_i p_i |e_i\rangle \langle{e_i}| e_j = \sum_i p_i |e_i\rangle \delta^i_j = p_j |e_j\rangle $$ I would write, assuming $|\psi\rangle = |e_j\rangle$: $$ O |\psi\rangle= O |e_j\rangle=\sum_i p_i |e_i\rangle \langle{e_i}| e_j\rangle = \sum_i p_i |e_i\rangle \delta^i_j = p_j |e_j\rangle $$

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