# Physical interpretation of the bra-ket notation

The bra-ket notation generally consists of 'ket', i.e. a vector, and a 'bra', i.e. some linear map that maps a vector to a number in the complex plane.

Now, using this bra-ket notation we can compute the inner product of some operator, say $$\hat{H}$$, so $$\langle\psi|\hat{H}|\psi\rangle$$ defines the eigenvalue of some hermitian operator $$\hat{H}$$. This is also called the expectation value of $$\hat{H}$$ and describes the probability of measuring this operator given the state $$\psi$$. I hope this is correctly understood.

We can also derive the inner product $$\langle\phi|\psi\rangle$$. I must admit that I am a little confused about this representation although it makes sence mathematically. Does this mean the probability of being in the state $$\phi$$ given the state $$\psi$$? I hope someone can clarify.

• $\left<\psi\mid\hat H\mid\psi\right>$ does NOT define an eigenvalue. Would it help you if you convert $\left<\phi\mid\psi\right>$ to $\left<\phi\mid\hat{𝟙}\mid\psi\right>$? Jul 5, 2023 at 17:06
• Multiplying a vector by the unit matrix yields a vector with all the diagonal components. How does this help me realize the physical meaning of $\langle\phi|\psi\rangle$? Jul 5, 2023 at 17:10
• because by your question statement it seemed like you think you understood the bra-op-ket notation, yet not the bra-ket notation. Just worth asking explicitly. Jul 5, 2023 at 17:22

Now, using this bra-ket notation we can compute the inner product of some operator, say $$\hat{H}$$, so $$\langle\psi|\hat{H}|\psi\rangle$$ defines the eigenvalue of some hermitian operator $$\hat{H}$$.

The inner product is a thing between two vectors - "the inner product of some operator" is not a meaningful phrase. If $$|\psi\rangle$$ is a normalized eigenvector of $$\hat H$$ with eigenvalue $$\lambda$$, then it's true that $$\langle \psi|\hat H|\psi\rangle = \lambda$$, but the definition of an eigenvector/eigenvalue pair is that $$\hat H|\psi\rangle = \lambda|\psi\rangle$$.

This is also called the expectation value of $$\hat{H}$$ and describes the probability of measuring this operator given the state $$\psi$$.

$$\langle \psi|\hat H|\psi\rangle$$ is referred to as the expectation value (or expected value) of $$\hat H$$ (corresponding to the normalized state vector $$|\psi\rangle$$). The interpretation of this number is that if you take a large number of identical systems all prepared in the state $$|\psi\rangle$$ and measured $$\hat H$$ in each of them, you would expect the mean value of all of those results to be $$\langle \psi|\hat H|\psi\rangle$$.

We can also derive the inner product $$\langle\phi|\psi\rangle$$. I must admit that I am a little confused about this representation although it makes sence mathematically. Does this mean the probability of being in the state $$\phi$$ given the state $$\psi$$?

There is no immediate physical interpretation of the inner product between two vectors - it is a quantity which shows up in all kinds of different contexts, and essentially measures the "overlap" between $$\psi$$ and $$\phi$$. It is analogous to the ordinary dot product between vectors in $$\mathbb R^3$$.

If $$\psi$$ is a normalized state vector representing the state of the system and $$\phi$$ is a normalized eigenvector of some observable $$\hat A$$ with (non-degenerate) eigenvalue $$\lambda$$, then $$|\langle \phi|\psi\rangle|^2$$ is the probability of measuring $$\hat A$$ to take the value $$\lambda$$. So that is one context in which the expression could arise. But trying to assign a single physical meaning to the inner product is like trying to assign a single physical meaning to the dot product between vectors in $$\mathbb R^3$$.

• Thank you very much for the answer. It was very informative and satisfying. Jul 5, 2023 at 19:16

If $$H$$ is a Hermitian operator, then $$\langle \psi|H|\psi\rangle$$ represents the expectation value (essentially "average measurement result") when the observable associated with $$H$$ is measured. Suppose $$H = \pmatrix{0 \ 1 \\ 1 \ 0}$$ and $$|\psi\rangle = \pmatrix{1 \\ 0}$$. If a measurement of the observable associated with $$H$$ is done on the state $$\psi$$, the expectation value is 0 ($$\langle\psi|H|\psi\rangle = \pmatrix{1 \ 0}\pmatrix{0 \\ 1} = 0$$). This isn't in any sense an eigenvalue of $$H$$, the two eigenvalues of $$H$$ are 1 and -1, and it is in fact impossible for a measurement of the observable associated with $$H$$ to result in a value of 0, but this still does have meaning that will be established.

When $$|\psi\rangle$$ undergoes a measurement associated with $$H$$'s observable, it will collapse to an eigenvector of $$H$$ and the measurement will result in the value of the associated eigenvector, and the two unit eigenvectors (up to multiplication by a complex unit circle scalar) of $$H$$ in the above example are $$\pmatrix{1/\sqrt{2} \\ 1/\sqrt{2}}$$ and $$\pmatrix{1/\sqrt{2} \\ -1/\sqrt{2}}$$ respectively for 1 and -1. The Born rule states that the probability of $$|\psi\rangle$$ collapsing to another state $$|\phi\rangle$$ in this case is given as $$\langle\psi|\phi\rangle\langle\phi|\psi\rangle = |\langle\phi|\psi\rangle|^2$$ (since reversing the bra ket is equivalent to complex conjugation and $$a a^* = |a|^2$$). For the eigenvectors of $$H$$ and with $$|\psi\rangle = \pmatrix{1 \\ 0}$$, this value is equal to $$1/2$$ in both cases, so there is a half chance that, upon measuring $$H$$'s observable, the result will be 1, and a half chance the result will be -1. If this is attempted multiple times you will get an averaged result of 0, even though that is not a possible result of a single measurement, and that is what $$\langle\psi|H|\psi\rangle$$, the expectation value of a measurement on $$|\psi\rangle$$ of the observable associated with $$H$$, physically represents.

Now, $$|\phi\rangle\langle\phi|$$ is itself a Hermitian operator, and this means the Born rule probability value $$\langle\psi|\phi\rangle\langle\phi|\psi\rangle$$ can itself be interpreted as the expectation value of a measurement (one assumes all vectors here are normalized). This Hermitian operator has $$|\phi\rangle$$ as an eigenvector with eigenvalue 1 and all orthogonal states an eigenvalue of 0, meaning essentially a measurement of 1 corresponds to the state being measured as $$|\phi\rangle$$ and 0 if measured otherwise (but not probing any more information about the state other than it being orthogonal from $$|\phi\rangle$$). If the probability of measuring a 1 is $$p$$ and the other option is 0, then the average result will be $$p$$, so this interpretation still works.

So, $$\langle\phi|\psi\rangle$$ by itself is not a probability (for one it is complex), but it does, when multiplied by its complex conjugate or has its absolute value squared, correspond to the probability $$|\psi\rangle$$ will become $$|\phi\rangle$$ when a measurement that distinguishes $$|\phi\rangle$$ is performed. Note that in some higher dimensional cases $$|\phi\rangle$$ may be an eigenvector of an observable's operator but not the only one of that eigenvalue, and in that case a measurement of $$|\psi\rangle$$ that resulted in $$|\phi\rangle$$'s eigenvalue would collapse to an even projection of $$|\psi\rangle$$ onto the corresponding eigenspace rather than to $$|\phi\rangle$$. For example, if $$|\phi_1\rangle$$ and $$|\phi_2\rangle$$ were both eigenvectors of an operator $$A$$ with the same eigenvalue and all other eigenvectors of that eigenvalue are linear combinations of the two, the probability that value is measured on $$|\psi\rangle$$ is $$\langle \psi|(|\phi_1\rangle\langle \phi_1| + |\phi_2\rangle\langle \phi_2|)|\psi\rangle$$, and the state collapsed to will be $$|\phi_1\rangle\langle\phi_1|\psi\rangle + |\phi_2\rangle\langle\phi_2|\psi\rangle$$ normalized.

• Thank you very much for your answer. I need some time to let some of these things sink in. I will read the rest when I am ready to learn something new. But again, thank you. Jul 5, 2023 at 19:30

When first learning QM it is useful to write yourself a little linear algebra / quantum physics dictionary.

But first, the mathematical setup: one usually has a complex vector space* $$V$$ (usually a Hilbert space $$\mathcal{H}$$ or some technical extension) equipped with a hermitean form $$\langle{\cdot},\cdot\rangle$$. The latter gives us a canonical isomorphism between $$V$$ and the dual space $$V^*$$, defined by $$x \mapsto \phi_x$$ where $$\phi_x : y \mapsto \langle x,y \rangle$$. By linear operators we mean linear maps $$\mathcal{O}:V \rightarrow V$$. In QM, many important operators (in particular observables) are self-adjoint. Introductory books written by physicists sometimes use hats such has $$\hat{\mathcal{O}}$$ to distinguish $$\mathcal{O}$$ from its classical counterpart. I will forgo this since here the distinction is clear from context.

The beginning of our dictionary is:

• A ket $$|x\rangle$$ (the kets make up the physical states) $$\leftrightarrow$$ The element $$x \in V$$
• A bra $$\langle x |$$ $$\leftrightarrow$$ The element $$\phi_x \in V^*$$ as defined above.

Thus $$\langle x | y\rangle$$ is merely the hermitean product $$\phi_x(y) = \langle x , y \rangle$$.

For a self adjoint operator $$A$$, one has $$\langle x , A y \rangle = \langle Ax,y\rangle$$ and so one might as well write this as an ambivalent $$\langle x | A |y \rangle$$ in this situation. This is Dirac notation. It only makes sense when $$A$$ is self-adjoint. With this in mind, we can continue the dictionary:

• Oberservables $$\leftrightarrow$$ Self-adjoint operators $$A:V \rightarrow V$$
• The expectation value of an observable $$A$$ given a state $$|\psi\rangle$$ $$\leftrightarrow$$ $$E_A(\psi) = \langle \psi|A | \psi \rangle / \langle \psi | \psi \rangle$$.

There is a slight 'unexpected' notational advantage of Dirac notation, which is helpful in physics. This is that elements of $$V$$ (which in applications can be some messy tensor product) can be very easily notated. For example, labelling of hydrogen atom states is really clear with Dirac notation; e.g. $$|n,j,s\rangle$$ is easy to read whereas traditionally mathematicians would probably notate using subscripts $$v_{n,j,s}$$. You can even use words and symbols in kets, which would look horrible as subscripts.

I will stop here. You should continue to expand this dictionary as you learn more. In particular, try including eigenvalues/eigenvectors. Good books to learn from include those by Sakurai and Messiah.

*For the technically minded, states are really 'rays' in a vector space, or better, elements of the projective vector space $$PV$$. Doing calculations over $$PV$$ often amounts to going back out to $$V$$, but conceptually you should realize that $$PV$$ is the space you are really working over.

• Thank you. I think it is a great idea to couple these quantum mechanical concepts to concepts from linear algebra. I do have one question. Is it correctly understood that the expectation value of an observable, say $A$, can be written as either $\langle\psi|A|\psi\rangle$ or $\langle\psi|\psi\rangle$ Jul 5, 2023 at 21:40
• It's not really coupling these two concepts together. Linear algebra is the language of quantum mechanics. It seems that you might want a good course in linear algebra (at the level of Lang say). No. $\langle \psi | A | \psi \rangle$ is the object $\langle \psi, A \psi \rangle$ which obviously depends on $A$. This is what is meant by the expectation value of $A$. On the other hand, $\langle \psi | \psi \rangle$ has no $A$-dependence, indeed taking $\psi$ to be normalized by definition means $\langle \psi | \psi \rangle = 1$ trivially. Jul 5, 2023 at 21:54
• Thank you. Sakurai explains this concept really well also. I was just confused about the expectation value given as $E_A(\psi)=\langle\psi|A|\psi\rangle/\langle\psi|\psi\rangle$, but maybe the point is to normalize the expectation value? Jul 6, 2023 at 15:17
• Exactly. You could of course normalize $|\psi\rangle$ a priori in which case you don't need the denominator.. Jul 6, 2023 at 18:28